Class 12 Commerce Maths - Linear Programming

Solve the following equation
$\displaystyle 2\left ( 2x+3 \right )-10\leq 6\left ( x-2 \right )$

We have

$\displaystyle \Rightarrow 2(2x+3)-10\leq (x-2)$

$\displaystyle \Rightarrow 4x+6-10\leq 6x-12$

$\displaystyle \Rightarrow 4x-6x\leq 12+4$ [Transposing - 4 to RHS and 6x to LHS]

$\displaystyle \Rightarrow -2x\leq -8\: \: \Rightarrow \: \: \frac{-2x}{-2}\geq \frac{-8}{-2}$

$\displaystyle \Rightarrow x\geq 4\: \: \Rightarrow \: \: x\in [4,\infty )$
Hence the solution set of the given inequation is

$\displaystyle [4,\infty )$ which can be graphed on real line as shown in Figure
1032123_329132_ans_55654686308b4234ad62aeb4c481fff2.bmp

Check, whether the half plane $3x+6y \geq 0$ contains $(1, 1)$ , if so shade the plane containing $(1, 1)$ .

$3x+6y \geq 0$

but

$(1,1)\Rightarrow \boxed{9 \geq 0}$

True, hence

$(1,1)$ is contained in this reason.

1082513_515089_ans_dbf9958756a24382bd15088d9ee01484.png

Shade the following half plane: $5x+10y-20\geq 0$ .

$5x+10y-20 \geq 0$

1082556_515104_ans_5f4c3b768ba1452e9bdcb11d627f09da.png

Check, whether the half plane $x+y+4 > 0$ contains the origin. Identify the plane and shade it.

$x+y+4 > 0$

Put

$(0,0)\Rightarrow \boxed{4 > 0}$ (True)

Hence contain origin (: True)

1082500_515088_ans_596560b66a9e445ca832473d3f07fdfd.png

The solution set of constraints of Linear Programming problem is called

A feasible region or solution space is the set of all possible points problem that satisfy the problem's constraints, potentially including inequalities, equalities, and integer constraints.

What is meant by "Open Convex Region"?

A Convex set is a region such that, for every pair of points within the region, every point on the straight line segment that joins the pair of points is also within the region.

In the above figure, left one is Convex region and the right one is non-convex.

Define: 'Objective function'.

Objective function:- An equation to be optimized given certain constraints and with variables that need to be minimized or maximized using nonlinear programming techniques. An objective function can be the result of an attempt to express a business goal in mathematical terms for use in decision analysis, operations research or optimization.

Define feasible region in LPP.

A feasible region is an area defined by a set of coordinates that satisfy a system of inequalities. The region satisfies all restrictions imposed by a linear programming scenario. The concept is an optimization technique.

Determine graphically the minimum value of the objective function
$Z = 50x + 20y ... (1)$
subject to the constraints:
$2x y \geq 5 ... (2)$
$3x + y \geq 3 ... (3)$
$2x -3y \leq 12 ... (4)$
$x \geq 0, y \geq 0 ... (5)$ .

First of all, let us graph the feasible region of the system of inequalities

$(2)$ to

$(5)$ . The feasible region (shaded) is shown in the Fig. Observe that the feasible region is unbounded.
We now evaluate

$Z$ at the corner points.

Corner Point	$Z = -50x + 20y$
$(0, 5)$	$100$
$(0, 3)$	$60$
$(1, 0)$	$-50$
$(6, 0)$	$-300\leftarrow smallest$

From this table, we find that

$-300$ is the smallest value of

$Z$ at the corner point

$(6, 0)$ . Can we say that minimum value of

$Z$ is

$- 300$ ? Note that if the region would have been bounded, this smallest value of

$Z$ is the minimum value of

$Z$ (Theorem

$2$ ). But here we see that the feasible region is unbounded. Therefore,

$- 300$ may or may not be the minimum value of

$Z$ . To decide this issue, we graph the inequality

$- 50x + 20y < - 300$ (see Step

$3$ (ii) of corner Point Method.)
i.e.,

$-5x + 2y < - 30$
and check whether the resulting open half plane has points in common with feasible region or not.

If it has common points, then

$-300$ will not be the minimum value of

$Z$ .

Otherwise,

$-300$ will be the minimum value of

$Z$ .
As shown in the Fig, it has common points.

Therefore,

$Z = 50 x + 20 y$ has no minimum value subject to the given constraints.

Minimize $Z = 7x + y$ subject to
$5x + y\geq 5, x + y \geq 3, x \geq 0, y\geq 0$ .

The intersection point of

$5x+y=5$ and

$x+y=3$ is

$(0.5,2.5)$

The shaded region shows feasible area according to given constraints.

Value of

$Z$ at

$A=7\times 0+5=5$

Value of

$Z$ at

$B=7\times 3+0=21$

Value of

$Z$ at

$E=7\times 0.5+2.5=6$

Hence, minimum value of

$Z$ subject to given constraints is

$5$ .

Dietician has to develop a special diet using two foods $P$ and $Q$ . Each packet (containing $30\ g$ ) of food $P$ contains $12$ units of calcium, $4$ units of iron, $6$ units of cholesterol and $6$ units of vitamin A, while each packet of the same quality of food $Q$ contains $3$ units of calcium, $20$ units of vitamin A. The diet requires atleast $240$ units of calcium, atleast $460$ units of iron and almost $300$ units of cholesterol. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A?

Let

$x$ and

$y$ be the number of packets of food

$P$ and

$Q$ respectively.

Obviously

$x ≥ 0, y ≥ 0$ .

Mathematical formulation of the given problem is as follows:
Minimise

$Z = 6x + 3y$ (vitamin A)

subject to the constraints

$12x + 3y ≥240$ (constraint on calcium), i.e.

$4x + y ≥ 80$ ... (1)

$4x + 20y ≥460$ (constraint on iron), i.e.

$x + 5y ≥ 115$ ... (2)

$6x + 4y ≤300$ (constraint on cholesterol), i.e.

$3x + 2y ≤ 150$ ... (3)

$x ≥ 0, y ≥ 0$ ... (4)

$Z$ is minimum at the point

$(15, 20)$ .

Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if

$15$ packets of food

$P$ and

$40$ packets of food

$Q$ are used in the special diet.

The minimum amount of vitamin A will be

$285$ units.

A factory manufactures two types of screws $A$ and $B$ , each type requiring the use of two machines, an automatic and a hand-operated. It takes $4$ minutes on the automatic and $6$ minutes on the hand-operated machines to manufacture a packet of screws $'A'$ while it takes $6$ minutes on the automatic and $3$ minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most $4$ hours on any day. The manufacturer can sell a packet of screws $'A'$ at a profit of $70$ paise and screws $'B'$ at a profit of $Rs. 1$ . Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the given LLP and solve it graphically and find the maximum profit.

Let

$x$ be the number of packages of screws A and

$y$ be the number of packages of screws B that we can make.

Clearly,

$x, y\geq 0$ .

Total time on machine is

$4\ hr = 240\ min$

$\therefore$ for automatic machine , time =

$4x+6y\leq 240\implies 2x+3y\leq 120$

$\therefore$ for hand machine, time =

$6x+3y\leq 240\implies 2x+y\leq 80$

The shaded region is the feasible region.

The profits on Screw A is Rs. 0.7 and on Screw B is Rs. 1. We need to maximize the profits, i.e. maximize

$z=0.7x + y$ , given the above constraints.

Now, at A,

$z=28$

at O,

$z=0$

at E,

$z=41$

at D,

$z=40$

Hence, maximum profit is at point

$E(30,20)$ .

Minimise $Z = 3x + 2y$
subject to the constraints:
$x + y \geq 8 ... (1)$
$3x + 5y \leq 15 ... (2)$
$x \geq 0, y \geq 0 ... (3)$ .

Let us graph the inequalities

(1)

(1) to

(3)(3) (Fig).
From Fig, you can see that there is no point satisfying all the constraints simultaneously. Thus, the problem is having no feasible region and hence no feasible solution.

Solve the following inequalities by graphical method:
$2x+y\ge 6,3x+4y\le 12$

Solve for the points where the lines intersect the coordinate axes:

$2x+y\geq 6 \Rightarrow \dfrac{x}{3} + \dfrac{y}{6} \geq 1 \Rightarrow A = (3,0), \ B = (0,6)$

$3x+4y\leq 12 \Rightarrow \dfrac{x}{4} + \dfrac{y}{3} \leq 1 \Rightarrow C = (4,0), \ D = (0,3)$

The lines intersect at

$E = (12/5,6/5)$ .

Solving the inequalities we get

$x\ge \dfrac{12}{5}$ and

$y\le \dfrac{6}{5}$

The feasible region is the triangular region

$\Delta AEC$ since

$x\ge \dfrac{12}{5}$ and

$y\le \dfrac{6}{5}$ .

Solve the following linear programming problem graphically:
Maximise $Z = 4x + y ... (1)$
subject to the constraints: $x + y \leq 50 ... (2)$
$3x + y\leq 90 ... (3)$
$x\geq 0, y\geq 0 ... (4)$ .

The shaded region in fig. is the feasible region determined by the system of constraints

$(2)$ to

$(4)$ . We observe that the feasible region

$OABC$ is bounded. So,we now use Corner Point Method to determine the maximum value of

$Z$ .
The coordinates of the corner points

$O, A, B$ and

$C$ are

$(0, 0), (30, 0), (20, 30)$ and

$(0, 50)$ respectively. Now we evaluate

$Z$ at each corner point.

Corner Point	Corresponding value of $Z$
$(0, 0)$	$0$
$(30, 0)$	$120\leftarrow Maximum$
$(20, 30)$	$110$
$(0, 50)$	$50$

Hence, maximum value of

$Z$ is

$120$ at the point

$(30, 0)$ .

Find graphical solution for the following system of linear inequations:
$3x + 2y\leq 180; x + 2y \leq 120, x \geq 0, y\geq 0$
Hence find co-ordinates of corner points of the common region.

So, from graph corners points are :

(0,0),(0,60),(60,0),(30,45)

1163916_979551_ans_6c24b26312144814b10eeab5b295a431.png

(Diet problem) A dietician has to develop a special diet using two foods $P$ and $Q$ . Each packet (containing $30\ g$ ) of food $P$ contains $12$ units of calcium, $4$ units of iron, $6$ units of cholesterol and $6$ units of vitamin A. Each packet of the same quantity of food $Q$ contains $3$ units of calcium, $20$ units of iron, $4$ units of cholesterol and $3$ units of vitamin A. The diet requires atleast $240$ units of calcium, atleast $460$ units of iron and at most $300$ units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?

Let

$x$ and

$y$ be the number of packets of food

$P$ and

$Q$ respectively. Obviously

$x \geq 0, y \geq 0$ . Mathematical formulation of the given problem is as follows :
Minimise

$Z = 6x + 3y$ (vitamin A)
subject to the constraints

$12x + 3y \geq 240$ (constraint on calcium), i.e.

$4x + y \geq 80 ... (1)$

$4x + 20y \geq 460$ (constraint on iron), i.e.

$x + 5y \geq 115 ... (2)$

$6x + 4y \leq 300$ (constraint on cholesterol), i.e.

$3x + 2y \leq 150 ... (3)$

$x \geq 0, y \geq 0 ... (4)$
Let us graph the inequalities

$(1)$ to

$(4)$ .
The feasible region (shaded) determined by the constraints

$(1)$ to

$(4)$ is shown in Fig and note that it is bounded.
The coordinates of the corner points

$L, M$ and

$N$ are

$(2, 72), (15, 20)$ and

$(40, 15)$ respectively. Let us evaluate

$Z$ at these points:

Corner Point	$Z = 6x + 3y$
$(2, 72)$	$228$
$(15, 20)$	$150\leftarrow Minimum$
$(40, 15)$	$285$

From the table, we find that

$Z$ is minimum at the point

$(15, 20)$ . Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if

$15$ packets of food

$P$ and

$20$ packets of food

$Q$ are used in the special diet. The minimum amount of vitamin A will be

$150$ units.

Solve the following Linear Programming Problems graphically:
Maximise $Z = 3x + 2y$
subject to $x + 2y \leq 10, 3x + y \leq 15, x, y \geq 0$ .

The inequalities

$x,y\geq 0$ can be graphed.

By alternately putting x and y = 0 we can plot

$x + 2y =10$ and

$3x +y =15$

The inequalities can be graphed as shown.

We get polygon ABCD

We need to check value of Z at all 4 corner points.

$Z = 3x+2y$ must be maximised.

$Z(0,0) = 0; Z(5,0) = 15; Z(4,3) = 18; Z(0,5) = 10$

Clearly Z is maximised at (4,3) subject to given constraints.

Use graphical method
Maximize: $z=2x+3y$
Subject to:
$x+2y\le 40$
$6x+5y\le 150$
$x\ge 0, y\ge 0$

The feasible area is defined by the constraints as shown in the figure below in figure fig.3.4

Suppose that in addition to the existing constraints, the company is contracted to produce at least

$30$ units each week. This additional constraint can be written as :

$x+2y\ge 30$ . As a boundary solution, the constraint would be:

$x+y=30,(x=0,y=30)(x=30,y=0)$ . the three structural constraints are shown in the figure in fig. 3.5.

This case presents the manager with demands which cannot simultaneously be satisfied

1036042_849048_ans_003313001455472fa6aa7c5170715548.PNG

If possible, maximize $z = x + 4y$ , subject to $3x + 6y \leq 6, 4x + 8y \geq 16$ and $x \geq 0, y \geq 0$ .

Given:

$z = x + 4y$ , subject to

$3x + 6y \leq 6, 4x + 8y \geq 16$ and

$x \geq 0, y \geq 0$

After plotting

$3x + 6y \leq 6$ and

$4x + 8y \geq 16$ on graph paper we get the as shown.

As we can see in the image, there is no intersection area between the two equation, so it is

$not$

$possible$ to maximize

$z$ .

A paper mill produces rolls of paper used in making cash registers.Each roll of paper is $100$ m in length and can be used in widths of $3,4,6$ and $10$ cm.The company's production process results in rolls that are $24$ cm in width.Thus, the company must cut its $24$ cm roll to the desired widths.It has six basic cutting alternatives as follows:
Cutting
alternatives Width of Rolls(cm)
$3$ Width of Rolls(cm)
$4$ Width of Rolls(cm)
$6$ Width of Rolls(cm)
$10$ Waste(cm)
$1$ $4$ $3$ - - -
$2$ - $3$ $2$ - -
$3$ $1$ $1$ $1$ $1$ $1$
$4$ - - $2$ $1$ $2$
$5$ - $4$ $1$ - $2$
$6$ $3$ $2$ $1$ - $1$
The minimum demand for the four rolls is as follows:
Roll width(cm) Demand
$2$ $2,000$
$4$ $3,600$
$6$ $1,600$
$10$ $500$
The paper mill wishes to minimize the waste resulting from trimming to size.Formulate the L.P model.

Formulation of L.P model:
Key decision is to determine how the paper rolls be cut to the required widths so that trim losses(wastage) are minimum.
Let

${x}_{j}\left(j=1,2,...6\right)$ represent the number of times each cutting alternative is to be used.These alternatives result/do not result in certain trim loss.
Objective is to minimize the trim losses.
i.e., minimize

$Z={x}_{3}+2{x}_{4}+2{x}_{5}+{x}_{6}.$
Constraints are on the market demand for each type of roll width:
For roll width of

$3$ cm,

$4{x}_{1}+{x}_{3}+3{x}_{6}\ge2,000$
For roll width of

$4$ cm,

$3{x}_{1}+3{x}_{2}+{x}_{3}+4{x}_{5}+2{x}_{6}\ge3,600$
and For roll width of

$10$ cm,

${x}_{3}+{x}_{4}\ge500$
Since the variables represent the number of times each alternative is to be used,they cannot have negative values.

$\therefore {x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6},$ each

$\ge0$

(Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $8$ units of vitamin A and $10$ units of vitamin C. Food I contains $2$ units/kg of vitamin A and $1$ unit/kg of vitamin C. Food II contains $1$ unit/kg of vitamin A and $2$ units/kg of vitamin C. It costs $Rs 50$ per kg to purchase Food I and $Rs. 70$ per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.

Let the mixture contain

$x$ kg of Food I and

$y$ kg of Food II. Clearly,

$x \geq 0,y \geq 0$ . We make the following table from the given data:

Resources	Food I $(x)$	Food II $(y)$	Requirement
Vitamins A (units/ kg)	$2$	$1$	$8$
Vitaminc C (units/ kg)	$1$	$2$	$10$
Cost (Rs/ kg)	$50$	$70$

Since the mixture must contain at least

$8$ units of vitamin A and

$10$ units of vitamin C, we have the constraints:

$2x + y \geq 8$

$x + 2y \geq 10$
Total cost

$Z$ of purchasing

$x$ kg of food I and

$y$ kg of Food II is

$Z = 50x 70y$
Hence, the mathematical formulation of the problem is:
Minimise

$Z = 50x + 70y ... (1)$
subject to the constraints :

$2x + y \geq 8 ... (2)$

$x + 2y \geq 10 ... (3)$

$x, y \geq 0 ... (4)$
Let us the graph the inequalities

$(2)$ to

$(4)$ . The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.
Let us evaluate

$Z$ at the corner points

$A(0,8), B(2,4)$ and

$C(10,0)$ .

Corner Point	$Z = 50x + 70y$
$(0, 8)$	$560$
$(2, 4)$	$380\leftarrow Minimum$
$(10, 0)$	$500$

In the table, we find that smallest value of

$Z$ is

$380$ at the point

$(2,4)$ . Can we say that the minimum value of

$Z$ is

$380$ ? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality

$50x + 70y < 380$ i.e.,

$5x + 7y < 38$
to check whether the resulting open half plane has any point common with the feasible region. From the Fig, we see that it has no points in common.
Thus, the minimum value of

$Z$ is

$380$ attained at the point

$(2, 4)$ . Hence, the optimal mixing strategy for the dietician would be to mix

$2$ kg of Food I and

$4\ kg$ of Food II,and with this strategy, the minimum cost of the mixture will be

$Rs. 380$ .

Solve the following linear programming problem using graphical method
Minimize: $Z=60x+30y$
Subject to:
$2x+3y\ge 120$
$2x+y\ge 80$
$x\ge 0, y\ge 0$

corner points

$A=(0,80)$ and

$C=(60,0)$ are found by inspection
Point B:
System:

$2x+3y=120.....(1)$

$2x+y=80.........(2)$

$(1)-(2)=2y=40$

$y=20$
Substitute for

$y=20$ in (2):

$2x+20=80$

$2x=60$

$x=30$
Point

$B:(30,20)$
Extreme Values

Corner point	$Objective function$ Z=60x+30y
$(0,80)$	$60(0)+30(80)=2400$
$(30,20)$	$60(30)+30(20)=2400$
$(60,0)$	$60(60)+30(0)=3600$

From the table above, there are two minimum values for the objective function:

$A=(0,80)$ and

$B=(30,20)$ . In this situation, the objective function will have the same minimum value

$(2,400)$ at all points the boundary line segment A and B

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A requires $5$ minutes each for cutting and $10$ minutes each for assembling. Souvenirs of type B requires $8$ minutes each for cutting and $8$ minutes each for assembling. There are $3$ hours $20$ minutes available for cutting and hours for assembling. The profit is Rs. $5$ each for type A and Rs. $6$ each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

1327411_1082953_ans_f75daf647f9c43538f05f762343c890b.jpg

Find the maximum value of $4x+7y$ with the conditions $3x+8y\leq 24, y\leq2, x\geq 0$ and $y\geq0$ .

Given condition

$3x+8y \leq 24$

$y\leq 2, x\geq0, y\geq0$ .

The vertices of the feasible region are

$(0,0), (8,0), \left(\dfrac{8}{3},0\right) and \, (0,2).$

The maximum value of the objective function attains at

$(8,0).$

$f=4x+7y$

The maximum value =

$4\times 8+7\times 0=32$ .

1327278_1053582_ans_01eaaa16de624d748fcd95b7b13d04f8.png

A library has to accommodate two different types of books on a shelf. The books are $6$ cm and $4$ cm thick and weight $1$ kg and $1\dfrac{1}{2}$ kg each respectively. The shelf is $96$ cm long and atmost can support a weight of $21$ kg. How should the shelf be filled with the books of two types in order to include greatest number of books. Make it a linear programming problem (LPP) .

Let two types of books be

$x$ and

$y$ respectively, Then the required LPP is

Max

$Z= x+y$

Subject to the constraints,

$6x+4y\le 96$ or

$3x+2y\le 48$ and

$x+\dfrac{3}{2}y\le 21$ or

$2x+3y\le 42$ with

$x,y\ge 0$ .

The taxi fare in a city is as follows: For the first kilometre, the fare is $Rs 8$ and for the subsequence distance it is $Rs 5 per Km$ . Taking the distance covered as $x Km$ and total fare as $Rs y.$ write linear equation for this information, and draw its graph.

x	0	1	2
y	3	8	13

Given, the total distance covered

$= x\ km=1+(x-1)\ km$

Fare For first kilometre

$= 8$

Fare for subsequent distance

$=5\ per\ km$

Fare for next

$(x-1)km= 5(x-1)$

Total fare

$= y$

$8+5(x-1)=y$

$8+5x-5=y$

$5x-y+3=0$

Which is the required linear equation.

$y= 5x+3$

$x = 0$ ,then

$y = 3$ , at

$x=1,$ then

$y=5+3=8$

$x= 2,$ then

$y= 10+3=13$ and so on.

Hence, this is the answer.

1034383_1091028_ans_629f701f1dba40f6a618f968351ea945.png

Formulate this problem an LLP and solve it graphically.
A Diet for a sick person must contain at least $4000$ units of vitamins, $50$ units of minerals and $1400$ units of calories. Two foods $A$ and $B$ are available at a cost of Rs $5$ ans Rs $4$ per unit respectively. one unit of food $A$ contains $200$ units of vitamins, $1$ units of minerals and $40$ units of calories, while one unit of the food $B$ contains $100$ unit of vitamins, $2$ unit of minerals and $40$ units of calories. Find what combination of the foods $A$ and $B$ should be used to have least cost, but it must satisfy the requirements of the sick person.. form the question as LLP and solve it graphically.

We have,

Let $x$ units of food A

$y$ units of food B

Where $x\ge 0,\,y\ge 0$

Then table

We are to minimize

$z=5x+4y$

Subject to constraints

$200x+100y\ge 4000$

$2x+y\ge 40\,\,\,\,......\,\,\left( 1 \right)$

$x+2y\ge 50\,\,\,\,......\,\,\left( 2 \right)$

$40x+40y\ge 1400$

$x+y\ge 35\,\,\,\,.......\,\,\left( 3 \right)$

$\text{where}\,\,x\ge 0,\,y\ge 0$

From equation (1) to and we get,

$2x+y=40$

If $x=0$ then $y=40$ OY in $L\left(0,40 \right)$

If $y=0$ then $x=20$ OX in $A\left( 20,0\right)$

From equation (2) to and we get,

$x+2y=50$

If $x=0$ then $y=25$ OY in $M\left( 0,25 \right)$

If $y=0$ then $x=50$ OX in $B\left( 50,0 \right)$

From equation (3) to and we get,

$x+y=35$

If $x=0$ then $y=35$ OY in $N\left( 0,35 \right)$

If $y=0$ then $x=35$ OX in $C\left( 35,0 \right)$

Shaded region is the feasible region, which is unbounded.

Then,

Corner point is

$B\left( 50,0 \right)$

$D\left( 20,15 \right)$

$E\left( 5,30 \right)$

$L\left( 0,40 \right)$

At point $B\left( 50,0 \right)\Rightarrow Z=5\left( 50 \right)+4\left( 0 \right)=250$

At point $D\left( 20,15 \right),\,Z=5\left( 20 \right)+4\left( 15 \right)=100+60=160$

At point $E\left( 5,\,30 \right),\,Z=5\left( 5 \right)+4\left( 30 \right)=25+120=145$

At point $L\left( 0,40 \right),\,Z=5\left( 0 \right)+4\left( 40 \right)=0+160=160$

Then at least cost $=Rs\,\,145\,at\,\left( 5,30 \right)$

Hence, this is the answer.
1207657_1147748_ans_01a4215cd5cd4739ab11ccff0d40c3fb.png

Solve each of the following linear programming problem by graphical method.
Maximize $Z=18x+10y$
Subject to $4x+y\le 20$
$2x+3y\le 30$
$x,y\ge 0$

1617747_1664144_ans_ce903f2332a74b81a9bf1b85e1f92677.jpg

Solve the following systems of inequalities graphically: $x+y\leq 9, y > x, x\geq 1$

Given:

$x+y\le 9$ y>x

$,x\ge 1$

Solution:Let us first convert into equalities

$x+y$ =9

$y=0$ if

$x=0$

then

$x-9$ then

$y=9$

(9,0) (0,9)

ii)

$y=x$

iii)

$x=1$

The shaded portion is the solution for the given three inequalities.

Solve $3x+2y\leq 12, x+2y\leq 16, x+y\leq 10, x\geq 0, y\geq 0.$

Given:

$3x+2y\le 12$

$x+2y\le 16$

$x\ge 0$

$y\ge 0$

Solution i)

$3x+2y=12$

$x$	0	6
$y$	4	0
Points	A(0,4)	B(6,0)

ii)

$x+2y=16$

$x$	0	16
$y$	8	0
Points	C(0,8)	D(16,0)

iii)

$x+y=10$

$x$	0	0
$y$	10	0
Points	E(0,10)	F(10,0)

The feasible region is AOB since it is common to the three equation graphyically with the points .

$A(0,4) , 0(0,0) and B(6,0)$ .

Suppose f is the collection of all ordered pairs of real numbers and x=6 is the first element of some ordered pair in f. Suppose the vertical line through x=6 intersects the graph of f twice. Is f a function? Why or why not?

No,

$f$ is not a function because for every value of first element of ordered pair there must be only one element of second element of an ordered pair related .

A housewife wishes to mix together two kinds of food, $X$ and $Y$ , in such a way that the mixture contains at least $10$ units of vitamin $A$ , $12$ units of vitamin $B$ and $8$ units of vitamin $C$ . The vitamin contents of one kg of food is given below :
Vitamin A Vitamin B Vitamin C
Food X 1 2 3
Food Y 2 2 1
One kg of food $X$ costs Rs. $6$ and one kg of food $Y$ costs Rs. $10$ . Formulate the above problem as a linear programming problem and find the least cost of the mixture which will produce the diet graphically. What value will you like to attach with this problem ?

Let

$x$ kg of food

$X$ and

$y$ kg of food

$Y$ are mixed.
Let

$z$ represents the cost function.
Min

$z=6x+10y$
According to LPP,

$x+2y\ge 10$

$2x+2y\ge 12 \Rightarrow x+y \ge 6$

$3x+y\ge 8$

We get two points in the solution set,

$(1,5)$ and

$(2,4)$
At

$(1,5)$ ,

$z=6+50=56$
At

$(2,4)$ ,

$z=12+40=52$
Hence, least cost of the mixture produced when

$2$ kg of food

$X$ and

$4$ kg of food

$Y$ are mixed.

Draw the line passing through(2, 3) and (3,2). Find the sum of coordinates of the points at which this line meets the x-axis and y-axis.

1958132_301836_ans_f7a8b1c876e9431393cd1e8d73b61541.jpg

Solve $4x + 3y > 12$ graphically

Graph of

$4x + 3y > 12$ is given as dotted line in the fig
This line divides the xy plane in two half planes I and II We select a point (not on the line) say (0, 0) which lies in one of the half planes and determine if this point satisfies the given inequality we note that

$4(0) + 3(0) > 12$
or

$0 > 12$ which is false
Hence half plane I is not the solution region of the given inequality Clearly any point on the line does not satisfy the given strict inequality In other words the shaded half plane II excluding the points on the line is the solution region of the inequality
1031688_311091_ans_75e8a848f84c42ee87ebd7a115f48b24.bmp

1031688_311091_ans_75e8a848f84c42ee87ebd7a115f48b24.bmp

Solve the following inequations $\displaystyle \frac{2x+4}{x-1}\geq 5$

We have

$\displaystyle \frac{2x+4}{x-1}\geq 5\Rightarrow \frac{2x+4}{x-1}-5\geq 0$

$\displaystyle \Rightarrow \: \: \: \frac{2x+4-5(x-1)}{x-1}\geq 0\: \: \: \Rightarrow \: \frac{2x+4-5x+5}{x-1}\geq 0$

$\displaystyle \frac{-3x+9}{x-1}\geq 0$ [Multiplying both sides by -1]

$\displaystyle \frac{3x-9}{x-1}\leq 0\: \: \Rightarrow \: \: \frac{3(x-3)}{x-1}\leq 0$ [Dividing both sides by 3]

$\displaystyle \frac{x-3}{x-1}\leq 0\: \: \Rightarrow \: \: 1<x\leq 3\Rightarrow x\in (1,3]$
Hence the solution set of the given inequations is (1, 3]
1032148_329145_ans_50d40c92443d4af8a740ce6cd5eb2740.bmp

1032148_329145_ans_50d40c92443d4af8a740ce6cd5eb2740.bmp

Solve & graph the solution set of $-2 < 2x - 4$ and $\displaystyle -2x+5\geq 13,x\: \epsilon \: R$

$-2 < 2x - 4$

$\displaystyle \Rightarrow 2x-4>-2\: and\: -2x+5\geq 13$

$\displaystyle \Rightarrow 2x>2\: and\: -2x\geq 13-5$

$\displaystyle \Rightarrow x>1\: \: \: and\: \: \: -2x\geq 8$

$\displaystyle \Rightarrow x>1\: \: \: \: and\: \: \: \: -x\geq 4$

$\displaystyle \Rightarrow x>1\: \: \: \: and\: \: \: \: x\leq -4$

$\displaystyle \therefore x>1\: or\: x\leq -4\: or\: x\in (-\infty ,-4)\cap (1,\infty )$
1032112_329130_ans_78b03ff9dff54d54bb1d58280ab209f4.bmp

A factory makes tennis rackets and cricket bats. A tennis racket takes $1.5$ hours of machine time and $3$ hours of craftsman's time in its making while a cricket bat takes $3$ hour of machine time and $1$ hour of craftsman's time. In a day, the factory has the availability of not more than $42$ hours of machine time and $24$ hours of craftsman's time.

If the profit on a racket and on a bat is Rs. $20$ and Rs. $10$ respectively, find the maximum profit of the factory when it works at full capacity

The given information can be compiled in a table as follows.

	Tennis Racket	Cricket Bat	Availability
Machine Time (h)	$1.5$	$3$	$4.2$
Craftsman's Time (h)	$3$	$1$	$24$

$\therefore$

$1.5x+3y\le 42$

$3x+y\le 24$

$x,y\ge 0$
The profit on a racket is Rs.

$20$ and on a bat is Rs.

$10$

$\therefore$

$Z=20x+10y$
The mathematical formulation of the given problem is
Maximise

$Z=20x+10y.......(1)$
subject to the constraints

$1.5x+3y \le 42..........(2)$

$3x+y\le 24............(3)$

$x,y\ge 0..........(4)$

The feasible region determined by the system of constraints is as shown.

The corner points are

$A(8,0), B(4,12), C(0,14)$ and

$0(0,0).$

The values of Z at these corner points are tabulated as shown in above figure

REF. IMAGE 2

The maximum value of Z is 200, which occurs at

$x = 4$ and

$y = 12.$

Thus, the factory must produce

$4$ tennis rackets and

$12$ cricket bats to earn the maximum profit. The maximum profit earned by the factory will be Rs

$200.$

1521835_423011_ans_cfa394aca9834bf5bf91a4e16999fce4.png

Reshma wishes to mix two types of food $P$ and $Q$ in such a way that the vitamin contents of the mixture contain at least $8$ units of vitamin $A$ and $11$ units of vitamin $B$ . Food $P$ costs $Rs.\ 60/kg$ and Food $Q$ costs $Rs.\ 80/kg$ . Food $P$ contains $3\ units/kg$ of vitamin $A$ and $5\ units/kg$ of vitamin $B$ while food $Q$ contains $4\ units/kg$ of vitamin $A$ and $2\ units/kg$ of vitamin $B$ . Determine the minimum cost of the mixture.

Let the mixture contain

$x$ kg of food

$P$ and

$y$ kg of food

$Q$ . Therefore,

$x\ge 0$ and

$y\ge 0$
The given information can be compiled in a table as follows.

	Vitamin A (units/kg)	Vitamin B (units/kg)	Cost (Rs/kg)
Food P	$3$	$5$	$60$
Food Q	$4$	$2$	$80$
Requirement (units/kg)	$8$	$11$

The mixture must contain at least

$8$ units of vitamin A and

$11$ units of vitamin

$B$
Therefore, the constraints are

$3x+4y\ge 8$

$5x+2y\ge 11$
Total cost,

$Z$ of purchasing food is,

$Z=60x+80y$
The mathematical formulation of the given problem is
Minimise

$Z=60x+80y........(1)$
subject to the constraints

$3x+4y\ge 8......(2)$

$5x+2y\ge 11....(3)$

$x,y\ge 0......(4)$
The feasible region determined by the system of constraints is as follows.

Corner point	$Z=60x+80y$
$A(\cfrac{8}{3},0)$	$160$	$\rightarrow$ Minimum
$B(2,\cfrac{1}{2})$	$160$	$\rightarrow$ Minimum
$C(0,\cfrac{11}{2})$	$440$

As the feasible region is unbounded, therefore,

$160$ may or may not be the minimum value of

$Z$ .
For this, we graph the inequality,

$60x+80y< 160$ or

$3x+4y< 8$ and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with

$3x+4y< 8$
Therefore, the minimum cost of the mixture will be Rs.

$160$ at the line segment joining the points

$(\cfrac{8}{3},0)$ and

$(2,\cfrac{1}{2})$

Solve the following system of linear inequality graphically
$\displaystyle 2x+y\geq 6$ ......(1)
$\displaystyle x-y\leq 2$ ......(2)

The graph of linear equation

$2x + y = 6$ is drawn in fig We note that solution of inequality (1) is represented by the shaded region above the line

$2x + y = 6$ including the point on the line
On the same set of axes we draw graph of the equation

$x - y = 2$ as shown in the fig Then we not that inequality (2) represents the shaded region above the line

$x - y = 2$ including the points on the line
Clearly the double shaded region common to the above two shaded region is the required solution region of the given system if inequalities
1031695_311152_ans_24f284f5c5ce4f85847c34a0dc6b7060.bmp

1031695_311152_ans_24f284f5c5ce4f85847c34a0dc6b7060.bmp

A diet is to contain at least $80$ units of vitamin $A$ and $100$ units of minerals. Two foods ${F}_{1}$ and ${F}_{2}$ are available. Food ${F}_{1}$ costs Rs. $4$ per unit food and ${F}_{2}$ s costs Rs. $6$ per unit. One unit of food ${F}_{1}$ contains $3$ units of vitamin $A$ and $4$ units of minerals. One unit of food ${F}_{2}$ contains $6$ units of vitamin $A$ and $3$ units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the mineral nutritional requirements?

Let the diet contain

$x$ units of food

${F}_{1}$ and

$y$ units of food

${F}_{2}$ .

Therefore,

$x\ge 0$ and

$y\ge0$
The given information can be compiled in a table as follows

	Vitamin A(units)	Mineral (units) Cost Per unit	(Rs)
Food ${F}_{1}(x)$	$3$	$4$ $4$
Food ${F}_{2}(y)$	$6$	$3$ $6$
Requirement	$80$	$100$

The cost of food

${F}_{1}$ is Rs.

$4$ per unit and of food

${F}_{2}$ is Rs.

$6$ per unit. Therefore, the constraints are

$3x+6y\ge 80$

$4x+3y\ge 100$

$x,y\ge 0$
Total cost of the diet,

$Z=4x+6y$
The mathematical formulation of the given problem is
Minimise

$Z=4x+6y......(1)$
subject the constraints

$3x+6y\ge 80......(2)$

$4x+3y\ge 100.....(3)$

$x,y\ge 0.......(4)$
The feasible region determined by the constraints is as given.
It can be seen that the feasible region is unbounded
The corner points are

$A\left(\cfrac{80}{3},0\right),B\left(24,\cfrac{4}{3}\right)$ and

$C\left(0,\cfrac{100}{3}\right)$
The values of

$Z$ at these corner points are as follows.

Corner point	$Z=4x+6y$
$A\left(\cfrac{80}{3},0\right)$	$\cfrac{320}{3}=106.67$
$B\left(24,\cfrac{4}{3}\right)$	$104$	$\rightarrow$ Minimum
$C\left(0,\cfrac{100}{3}\right)$	$200$

As the feasible region is unbounded, therefore,

$104$ may or may not be the minimum value

$Z$
For this, we draw a graph of the inequality,

$4x+6y< 104$ or

$2x+3y< 52$ , and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with

$2x+3y< 52$
Therefore, the minimum cost of the mixture will be Rs.

$104$ .

A farmer mixes two brands $P$ and $Q$ of cattle feed. Brand $P$ , costing Rs. $250$ per bag contains $3$ units of nutritional element $A$ , $2.5$ units of element $B$ and $2$ units of element $C$ . Brand $Q$ costing Rs. $200$ per bag contains $1.5$ units of nutritional elements $A$ , $11.25$ units of element $B$ and $3$ units of element $C$ . The minimum requirements of nutrients $A, B$ and $C$ are $18$ units, $45$ units and $24$ units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag> What is minimum cost of the mixture per bag?

Let the farmer mix

$x$ bags of brand

$P$ and

$y$ bags of brand

$Q$ .
The given information can be compiled in a table as follows.

	Vitamin A (units/kg)	Vitamin B (units/kg)	Vitamin C (units/kg)
Food P	$3$	$5$	$60$
Food Q	$4$	$2$	$80$
Requirement (units/kg)	$8$	$11$

The given problem can be formulated as follows
Minimise

$z=250x+200y.........(1)$
subject to the constraints

$3x+1.5y\ge 18........(2)$

$2.5x+11.25y\ge 45......(3)$

$2x+3y\ge 24........(4)$

$x,y\ge 0..............(5)$
The feasible region determined by the system of constraints is as shown
The corner points of the feasible region are

$A(18,0), B(9,2), C(3,6)$ and

$D(0,12)$
The values of

$z$ at these corner points are as follows

Corner point	$x=250x+200y$
$A(18,0)$	$4500$
$B(9,2)$	$2650$
$C(3,6)$	$1950$	$\rightarrow$ Minimum
$D(0,12)$	$2400$

As the feasible region is unbounded, therefore,

$1950$ may or may not be the minimum value of

$z$ .
For this, we draw a graph of the inequality,

$250x+200y< 1950$ or

$5x+4y< 39$ and check whether the resulting half plane has points in common with the feasible region or not.
IT can be seen that the feasible region has no common point with

$5x+4y< 39$
Therefore, the minimum value of

$z$ is

$2000$ at

$(3,6)$
Thus,

$3$ bags of brand

$P$ and

$6$ bags of brand

$Q$ should be used in the mixture to minimize the cost to Rs.

$1950$

1038797_423241_ans_7a1a090f600b4d848360a0b634d5ea45.png

If $S$ is the solution region of the inequalities $\begin{cases} x+3y\le 18 \\ 2x-3y\le 9 \end{cases}$ and $(a,b) \in S$ , then the minimum possible value for $b$ is

If $(a, b)$ is a solution to the system, then $a$ is the $x$ -coordinate of any point in the darkest shaded region and $b$ is the corresponding $y$ -coordinate.

When $a = 6$ , the minimum possible value for $b$ lies on the lower boundary line, $2x - 3y \leq 9$ .

It looks like the $y$ -coordinate is $1$ , but to be sure, substitute $x = 6$ into the equation and solve for $y$ .

You can use $=$ in the equation, instead of the inequality symbol, because you are finding a point on the boundary line.

$2x – 3y = 9$

$\Rightarrow 2(6) – 3y = 9$

$\Rightarrow 12 – 3y = 9$

$\Rightarrow -3y = -3$

$\Rightarrow y = 1$

So, the value of $b$ is $1$ .

Solve the following systems of inequalities graphically: $x+y\leq 9, y>x, x\geq 1$ .

$x+y\leq 9 ; y >x; x\geq 1$

1082554_515105_ans_8d1de42e371148ad8d5d892839761eac.png

Maximize $f=4x-y$ , subject to the constraints
$7x+4y\le 28$ , $2y\le 7$ , $x\ge 0$ , $y\ge 0$ .

Converting inequations into equations we get,

$7x+4y=28$

$2y=7\Longrightarrow y=\dfrac72=3.5$

$x=0,y=0$

Consider,

$7x+4y=28$

$x$	$0$	$4$
$y$	$7$	$0$

The feasible region is

$OABC.$

The values of the objective functions are given below:

Points	Value of objective function, $4x-y$
$O(0,0)$	$4\times0-0=0$
$A(4,0)$	$4\times4-0=16$
$B(2,3.5)$	$4\times2-3.5=4.5$
$(0,3.5)$	$4\times0-3.5=-3.5$

Thus,

$f$ is maximum at

$A(4,0)$ i.e.

$f_{max}=16$

Maximize $f=3x+y$ subject to the constraints
$8x+5y\le 40$ ; $4x+3y\ge 12$ ; $x\ge 0$ and $y\ge 0$ .

Solution:

Converting inequalities into equations:

$8x+5y=40$

$x$	$0$	$8$
$y$	$5$	$0$

$4x+3y=12$

$x$	$0$	$4$
$y$	$3$	$0$

$(3,0):f(x)=3x+y=3\times3+0=9$

$(5,0):f(x)=3x+y=3\times5+0=15$

$(0,8):f(x)=3x+y=3\times0+8=8$

$(0,4):f(x)=3x+y=3\times0+4=4$

Thus, at

$(5,0),\quad f$ is maximum.

A firm has the cost function $C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50$ and demand function $x = 100 - p$ .
Formulate the total profit function $P$ in terms of $x$

Cost function is

$C(x) = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50$
Demand function is

$x = 100 - p\Rightarrow p = 100 - x$
Revenue function,

$R(x) = p.x = x (100 - x) = 100x - x^{2}$

Profit function,

$P(x) = Revenue - Cost$

$= R(x) - C(x)$

$= 100x - x^{2} - \dfrac {x^{3}}{3} + 7x^{2} - 111x - 50$

$= -\dfrac {x^{3}}{3} + 6x^{2} - 11x - 50$

By graphical method solve the following linear programming problem for maximization.
Objective function Z = 1000 x + 600 y
Constraints $x + y \leq 200$
$4x - y \leq 0$
$x \geq 20, x \geq 0, y \geq 0$ .

Function to maximize:	$Z = 1000x + 600y$
Constraints:	$x+y≤200$
	$4x-y≤0$
	$x≥20$
	$x \geq 0, y\geq 0$

Corner Points	Values of $Z = 1000x+600y$ at corner points
$(20,80)$	$68000$
$(20, 180)$	$110000$
$\textbf{(40, 160)}$	$\textbf{136000}\ \ \leftarrow$ Maximum

Hence, the maximum value of

$Z = 1000x+600y$ is

$136000$ .

A firm has the cost function $C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50$ and demand function $x = 100 - p$ .
Write the total revenue function in terms of $x$

Cost function is

$C(x) = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50$
Demand function is

$x = 100 - p\Rightarrow p = 100 - x$
Revenue function,

$R(x) = p.x = x (100 - x) = 100x - x^{2}$

A shopkeeper sells not more than $30$ shirts of each colour. Atleast twice as many white ones are sold as green ones. If the profit on each of the white be Rs. $20$ and that of green be Rs. $25$ ; then find out how many of each kind be sold to give him a maximum profit. (Graph need not be drawn)

Let the number of white shirts be sold be

$x$

Then number of green shirts sold

$=2x$

Profit on each green shirt

$=$ Rs.

$25$

Profit on each white shirt

$=$ Rs.

$20$

$\therefore$ Total profit

$=2x\times 25+x\times 20=70x$

Shopkeeper sells no more than

$30$ shirts. so

$x\leq30$ and

$2x\leq30$

$x\leq15$

Maximum profit

$=70\times 15=$ Rs.

$1050$

A sweet-shop makes gift packet of sweets by combining two special types of sweets $A$ and $B$ which weigh $7\ kg$ . Atleast $3\ kg$ of $A$ and no more than $5\ kg$ of $B$ should be used. The shop makes a profit of Rs. $15$ on $A$ and Rs. $20$ on $B$ per $kg$ . Determine the product mix so as to obtain maximum profit.

Solution:

Let amount of

$A$ type of sweets

$=x\ kg$

Let amount of

$B$ type of sweets

$=y\ kg$

$\therefore x\ge0, y\ge0$

$--------(I)$

In one gift packet both types of sweets are to be included but not more than

$7\ kg.$

$\therefore x+y\le7$

$------(II)$

$A$ type of sweets must be at least

$3\ kg.$

$\therefore x\ge3$

$------(III)$

$B$ type of sweets no more than

$5\ kg.$

$\therefore y\le5$

$--------(IV)$

Profit Rs.

$15$ on

$A$ and Rs.

$20$ on

$B$ per

$kg$ .

$\therefore$ Profit function

$f=15x+20y$

The corner points for the graph are

$(3,4),(3,0)$ and

$(7,0)$

$x$	$y$	$f=15x+20y$
$3$	$4$	$125$
$3$	$0$	$45$
$7$	$0$	$105$

Maximum Profit of Rs.

$125$ can be gained when the mixture contains

$3\ kg$ of

$A$ and

$4\ kg$ of

$B.$

a) A manufacturing company makes two models $A$ and $B$ of a product. Each piece of model $A$ requires $9$ labor hours for fabricating and $1$ labor hour for finishing. Each piece of model $B$ requires $12$ labor hours for finishing and $3$ labor hours for finishing. For fabricating and finishing, the maximum labor hours available are: $180$ and $30$ respectively. The company makes a profit of Rs. $8000$ on each piece of model $A$ and Rs. $12000$ on each piece of model $B$ . How many pieces of model $A$ and model $B$ should be manufactured per week to realize a maximum profit? What is the maximum profit per week?
b) Find the value of $K$ so that the function $f (x) = \begin{matrix} \begin{cases}Kx+1, & \text{if }x \le 5 \\ 3x-5, & \text{if }x \le 5\end{cases}\end{matrix}$ at $x=5$ is a continuous function.

$(a)$

Let

$x$ be the number of pieces of model

$A$ and

$y$ be the number of pieces of model

$B$ .

Let total profit be

$Z$

Then, Total profit

$=8000x+12000y$

$Z=8000x+12000y$ ..............

$(i)$

Now we have the following mathematical model for the given problem

Max

$Z=8000x+12000y$

Subject to constraints

$9x+12y\leq 180$ .......... (Fabricating constraints)

$3x+4y\leq 60$ ..........

$(ii)$

$x+3y\leq 30$ (Finishing constraints)........

$(iii)$

$x\geq 0, y\geq 0$ (Non-negative constraints) ..........

$(iv)$

The shaded region

$OABC$ determined by linear inequalities

$(ii)$ and

$(iv)$ shown in the figure

Let us evaluate the objective function

$Z$ at each point as shown below

$O(0,0),$

$Z=8000(0)+12000(0)=0$

$A(20,0),\ \ Z=8000(20)+12000\times 0=160000$ \

$B(12,6), \ \ Z=8000\times 20+12000\times 6=168000$

$C(0,10),\ \ Z=8000\times 0+12000\times 10=120000$

Maximum value of

$Z$ is

$168000$ at

$B(12,6)$

Hence, the company should produce

$12$ pieces of Model

$A$ and

$6$ pieces of Model

$B$ to get the maximum profit i.e.

$168000$ .

$(b)$

Given function

$f (x) = \begin{matrix} \begin{cases}Kx+1, & \text{if }x \geq 5 \\ 3x-5, & \text{if }x \le 5\end{cases}\end{matrix}$ is continuous at

$x=5$

$\implies \lim_{x\rightarrow 5^{-}} f(x)=\lim_{x\rightarrow 5^{+}} f(x)$

$\implies \lim_{x\rightarrow 5^{-}} kx+1=\lim_{x\rightarrow 5^{+}}3x-5$

$\implies 5k+1=15-5$

$\implies k=\dfrac{9}{5}$

Solve the following L.P.P. by graphical method:

Maximise: $Z = 6x + 4y$

Subject to $x\leq 2, x + y \leq 3, -2x + y \leq 1, x \geq 0, y \geq 0$

To Maximize:	$Z = 6x+4y$
Constraints:	$x≤2$
	$x+y≤3$
	$-2x+y≤1$
	$x≥0,y≥0$

Plotting the constraints on the graph, we get the following points.

Points	$Z = 6x+4y$
$O(0,0)$	$0\ \ \leftarrow$ Minimum
$A(0,1)$	$4$
$B\left(\cfrac23,\cfrac73\right)$	$\cfrac{40}3$
$C(2,1)$	$16\ \ \leftarrow$ Maximum
$D(2,0)$	$12$

Hence, at

$C\equiv (2,1)$ , Maximum value of

$Z=6x+4y = 16$ .

Solve the following linear programming problem graphically:
Maximize $Z = 7x + 10y$
subject to the constraints
$4x + 6y \leq 240$
$6x + 3y \leq 240$
$x \geq 10$
$x \geq 0, y \geq 0$

The LPP given is:

Maximize

$7x+10y$ subject to contsraints

$4x+6y\leq 240\Rightarrow 2x+3y\leq120$

$6x+3y\leq240\Rightarrow2x+y\leq80$

$x\geq10$

$y\geq0$

Plot the graphs of

$2x+3y = 120, 2x+y=80, x=10$ . The shaded portion is the feasible region of the solution.

The corner points of the feasible region are

$(10,0),(40,0),(30,20)$ and

$\left(10,\dfrac{100}3\right)$

Examining the objective function at these values:

$(x,y)$	Value of objective function: $7x+10y$
$(10,0)$	$70$
$(40,0)$	$280$
$(30,20)$	$410$
$\left(10,\dfrac{100}3\right)$	$403.33$

From the values, we can see that

$7x+10y$ is maximum when

$x=30, y=20$ . The maximum value is

$410$ .

Consider the linear inequalities
$2x+3y\le 6, 2x+y\le 4, x\ge 0, y\ge 0$
(a) Mark the feasible region.
(b) Maximise the function $z=4x+5y$ subject to the given constraints.

Convert the linear inequalities into equalities to find the corner points

$2x+3y=6$ .........

$(i)$

$2x+y=4$ ...........

$(ii)$

From

$(i)$

$x=0\implies y=2$ and

$y=0$ when

$x=3$

So, the points

$(i)$ are

$(0,2)$ and

$(3,0)$ lie on the line given in

$(i)$

From

$(ii)$ , we get the points

$(0,4)$ and

$(2,0)$

Let's plot these point and we get the graph shown above.

The shaded part shows the feasible region.

The lines intersect at

$(\frac{3}{2},1)$ and other corner points of the region are

$(0,2), (2,0), (0,0)$ .

To maximize

$z$ , we need to find the value of

$z$ at the corner points

Corner points

$z=4x+5y$

$(0,0)$

$0$

$(2,0)$

$8$

$(0,2)$

$10$

$(\frac{3}{2},1)$

$11$

Thus,

$z$ is maximum at

$(\frac{3}{2}, 1)$

Solve graphically the linear inequalities
$2x+3y\le 7,x+2y\le 4$
$x\ge 0,y\ge 0$
If $z=6x+5y$ is the objective function, find its maximum value.

$z=6x+5y$

Convert the linear inequalities into equalities

$2x+3y=7$ .........

$(i)$

$x+2y=4$ ...........

$(ii)$

From

$(i)$

$x=0\implies y=2.3$ and

$y=0$ when

$x=3.5$

So, the points

$(0,2.3)$ and

$(3.5,0)$ lie on the line given in

$(i)$

From

$(ii)$ , we get the points

$(0,2)$ and

$(4,0)$

Let's plot these point and we get the graph shown above.

The shaded part shows the feasible region.

The lines intersect at

$(2,1)$ and other corner points of the region are

$(0,2), (3.5,0), (0,0)$ .

To find the maximum value of

$z$ , we need to find the value of

$z$ at the corner points

Corner points

$z=6x+5y$

$(0,0)$

$0$

$(3.5,0)$

$21$

$(0,2)$

$10$

$(2,1)$

$17$

Thus,

$z$ is maximum at

$(3.5,0)$ and its maximum value is

$21$ .

In a factory, there are two machines $A$ and $B$ producing toys. They respectively produce $60$ and $80$ units in one hour. $A$ can run a maximum of $10$ hours and $B$ a maximum of $7$ hours a day. The cost of their running per hour respectively amounts to $2,000$ and $2,500$ rupees. The total duration of working these machines cannot exceed $12$ hours a day. If the total cost cannot exceed Rs. $25,000$ per day and the total daily production is at least $800$ units, then formulate the problem mathematically.

Let

$x$ and

$y$ be the daily production units of machines

$A$ and

$B$ respectively.

The total daily production is atleast

$800$ .

$\Rightarrow x+y \geq 800$

$A$ can produce

$60$ units per hour for a maximum of

$10$ hours.

$\Rightarrow x \leq 60\times 10$

$\Rightarrow x \leq 600$

$B$ can produce

$80$ units per hour for a maximum of

$7$ hours.

$\Rightarrow y \leq 80\times 7$

$\Rightarrow y\leq 560$

The number of hours machine

$A$ runs is

$\dfrac{x}{60}$ and machine

$B$ runs is

$\dfrac{y}{80}$ .

Hence, total cost per day

$=$ Rs.

$\left(2000\dfrac{x}{60} + 2500\dfrac{y}{80}\right) =$ Rs.

$\left(\dfrac{100x}3 + \dfrac{125y}4\right) \leq$ Rs.

$25000$

Hence, the mathematical formulation of the given problem will be:

$\dfrac{100x}3+\dfrac{125y}4 \leq 25000$

$x+y\geq 800$

$x\leq 600$

$y\leq 560$

A furniture shop keeper deals only Teak and Rosewood tables which costs Rs. $25,000$ and Rs. $40,000$ respectively. His shop can contain a maximum of $250$ tables. The profit that he gets on these items are respectively Rs. $4,500$ and Rs. $5000$ .
He plans to invest a maximum of Rs. $70$ lakhs in his business.
(i) Formulate this problem.
(ii) Write the objective function if he wishes to make maximum profit.

Let

$x$ and

$y$ be the quantities of Teak and Rosewood tables respectively.

The shop can contain a maximum of

$250$ tables.

$\Rightarrow x+y \leq 250$

The shopkeeper can invest a maximum of Rs.

$70,00,000$

$\Rightarrow 25000x+40000y \leq 7000000$

$\Rightarrow 5x+8y\leq 1400$

Hence, the mathematical formulation of this problem is:

$x+y \leq 250$

$5x+8y \leq 1400$

Since the shopkeeper wishes to maximize his profit, the objective function will be to maximize

$Z: 4500x+5000y$ subject to the following conditions:

$x+y \leq 250$

$5x+8y \leq 1400$

Define feasible region.

A feasible region is an area defined by a set of coordinates that satisfy a system of inequalities. The region satisfies all restrictions imposed by a linear programming scenario. The concept is an optimization technique. For example, a planner can use linear programming to determine the best value obtainable under conditions dictated by several linear equations that relate to a real-life problem.

Consider the following L.P.P.
Maximize $Z=3x+2y$
Subject to the constraints
$x+2y\le 10$
$3x+y\le 15$
$x,y\ge 0$
(a) Draw its feasible region.
(b) Find the corner points of the feasible region.
(c) Find the maximum value of $Z$ .

(a)

$Z=3x+2y$ ..........

$(i)$

subject to the constraints

$x+2y\leq 10$

$3x+y\leq 15$

Convert these inequalities into equations

$x+2y=10$ .........

$(ii)$

$3x+y=15$ .........

$(iii)$

From

$(ii)$ , we get

$x=0\implies y=5$ and

$y=0$ when

$x=10$

So, the points

$(0,5)$ and

$(10,0)$ lie on the line given in

$(ii)$

From

$(iii)$ , we get the points

$(0,15)$ and

$(5,0)$

Let's plot these point and we get the graph in which, shaded part shows the feasible region.

(b)

Lines

$(ii)$ and

$(iii)$ intersect at

$(4,3)$ and other corner points of the region are

$(0,5), (5,0)$ and

$(0,0)$ .

(c)

To find the maximum value of

$z$ , we need to find the value of

$z$ at the corner points

Corner points

$z=3x+2y$

$(0,0)$

$0$

$(5,0)$

$15$

$(0,5)$

$10$

$(4,3)$

$18$

Thus,

$z$ is maximum at

$(4,3)$ and its maximum value is

$18$ .

a) Minimize and maximize $Z=x+2y$ , subject to the constraints
$x+2y \ge 100$
$2x-y \le 0$
$2x+y \le 200$
$x,y \ge 0$ by graphical method.
b) Prove that $\begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix}=4abc$

$(a)$

$Z=x+2y$ , subject to the constraints

$x+2y \ge 100$

$2x-y \le 0$

$2x+y \le 200$

$x-y \ge 0$ by graphical method.

On solving equations

$2x-y=0$ and

$x+2y=100$ we get point

$B(20,40)$

On solving

$2x-y=0$ and

$2x+y=200$ we get

$C(50,100)$

$\therefore$ Feasible region is shown by

$ABCDA$

The corner points of the feasible region are

$A(0,50), B(20,40), C(50,100), D(0,200)$

Let us evaluate the objective function

$Z$ at each corner points as shown below

$A(0,50),$

$Z=0+100=100$

$B(20,40),\ \ Z=20+80=100$

$C(50,100), \ \ Z=50+200=250$

$D(0.200),\ \ Z=0+400=400$

Hence, Maximum value of

$Z$ is

$400$ at

$D(0,200)$ and minimum value of

$Z$ is

$100$ at

$A$ and

$B$ .

$(b)$

Consider,

$\begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix}$

$=(b+c)\begin{vmatrix} c+a & b \\ c & a+b\end{vmatrix}-a\begin{vmatrix} b & b \\ c & a+b\end{vmatrix}+a\begin{vmatrix} b & c+a \\ c & c \end{vmatrix}$

$=(b+c)[(c+a)(a+b)-bc] - a[ab+b^2-bc] + a[bc-c^2-ac]$

$=(b+c)a(b+c+a)-a[b(b-c+a)]+a[-c(-b+c+a)]$

$=4abc$

Hence,

$\begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix}=4abc$

By Graphical method solve the following L.P.P under the following constraints.
$x+y\le 24,2x+y\le 32,x,y\ge 0$ . Find the maximum value of $z=150x+250y$

The shaded region is the feasible region.

The intersection point of

$2x+y=32$ and

$x+y=24$ is given as

$(8,16)$ .

Also,

$z=150x+250y=50(3x+5y)$

Now, considering the value of

$m=3x+5y$ at the extremums,

$(0,16)$ ,

$m=80$

$(24,0)$ ,

$m=72$

$(8,16)$ ,

$m=104$

$(0,0)$ ,

$m=0$

Hence, maxima of z occurs at

$(8,16)$ which is

$50m=5200$ .

A toy company manufactures two types of doll; a basic version-doll $A$ and a deluxe version doll $B$ . Each doll of type $B$ takes twice as long as to produce as one of type $A$ , and the company would have time to make a maximum of $2000$ per day if it produces only the basic version. The supply of plastic is sufficient to produce $1500$ dolls per day (both $A$ and $B$ combined). The deluxe version requires a fancy dress of which there are only $600$ per day available. If company makes profit of $Rs. 3$ and $Rs. 5$ per doll, respectively, on doll $A$ and $B$ ; how many each should be produced per day in order to maximize profit.

A farmer has a supply of chemical fertilizer of type $A$ which contains $10\%$ nitrogen and $6\%$ phosphoric acid and of type $B$ which contains $5\%$ nitrogen and $10\%$ phosphoric acid. After soil test, it is found that at least $7$ kg of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of type $A$ costs Rs. $5.00$ per kg and the type $B$ costs Rs. $8.00$ per kg. Using Linear programming, find how many kilograms of each type of the fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.

let the farmer bought $x$ kg of chemical $A$ and $y$ kg of $B$ , so

$x\ge 0.....(i)\\ y\ge 0.....(ii)$

Minimun nitrogen required $=7$

$\Rightarrow \dfrac { 10 }{ 100 } x+\dfrac { 5 }{ 100 }y \ge 7\\ \Rightarrow 2x+y\ge 140......(iii)$

Minimun phoshoric acid required $=7$ kg

$\Rightarrow \dfrac { 6 }{ 100 } x+\dfrac { 10 }{ 100 } y\ge 7\\ \Rightarrow 3x+5y\ge 350.....(iv)$

we have to minimise the cost

$\Rightarrow z=5x+8y$

from the graph critical points are $A(50,40),B(0,140)$ and $C(116.62,0)$

At $A$

$z=5(50)+8(40)=570$

At $B$

$z=0+8(140)=1120$

At $C$

$z=5(116.62)+0=583$

Cost is minimum at $A$

So the farmer must bought $50$ kg of chemical $A$ and $40$ kg of chemical $B$

Solve the linear programming problem by graphical method with the following restrictions.
$3x+5y \leq 5; 5x+2y\leq 10; x\geq 0, y\geq 0$ and maximize $Z=5x+3y$ .

From the graph it is very clear that dark shaded portion

$ACDE$ is the common (feasible) region for the given constraints.

From linear programming, we know that the maximum is obtained at the vertices of the feasible region.

So, lets find out

$Z=5x+3y$ values at the vertices,

$A (0,5) : Z = 5*0+3*5=15$

$C (0,1.68): Z=5*0+3*1.68 = 5.04$

$D(1.67,0): Z=5*1.67+3*0=8.35$

$E(2,0): Z=5*2+3*0=10$

$\therefore Z=15$ is the maximum at

$(0,5)$

Show the solution of the problem of linear programming under the following restrictions by graphical method: $x+y\leq 4, x \geq 0, y\geq 0$ .

Shaded region indicates of the graph, indicates the solution of the given linear programming problem.

A factory makes tennis rackets and cricket bats. A tennis racket takes $1.5$ hours of machine time and $3$ hours of craftmans time in its making while a cricket bat takes $3$ hour of machine time and $1$ hour of craftmans time. In a day, the factory has the availability of not more than $42$ hours of machine time and $24$ hours of craftsmans time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is $Rs. 20$ and $Rs. 10$ respectively, find the maximum profit of the factory when it works at full capacity.

Let number of tennis racket be made is

$X$ and number of cricket bat be made is

$Y$

Since, tennis bat requires

$1.5$ hours and cricket bat requires

$3$ hours of machine time. Also, there is maximum

$42$ hours of machine time available.

$\therefore 1.5X+3Y\leq 42$

$\Rightarrow X+2Y\leq 28\ \ \ ...(1)$

Since, tennis bat requires

$3$ hours and cricket bat requires

$1$ hours of craftmans time. Also, there is maximum

$24$ hours of craftmans time available.

$\therefore 3X+Y\leq 24\ \ \ ...(2)$

Since, count of an object can't be negative.

$\therefore X\geq 0,Y\geq 0\ \ \ ...(3)$

We have to maximize profit of the factory.

Here, profit on tennis racket is

$20\ Rs$ and on cricket bat is

$10\ Rs$

So, objective function is

$Z=20X+10Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=20X+10Y$
A $(0,14)$	$140$
B $(4,12)$	$200$ (maximum)
C $(8,0)$	$160$

Hence,

(i)

$4$ tennis rackets and

$12$ cricket bats must be made so that factory will work at full capacity.

(ii) Maximum profit of factory will be

$200\ Rs$

(Manufacturing problem) A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most $12$ hours whereas machine III must be operated for atleast $5$ hours a day. She produces only two items $M$ and $N$ each requiring the use of all the three machines.
The number of hours required for producing $1$ unit of each of $M$ and $N$ on the three machines are given in the following table:
Items Number of hours required on machines
I II III
$M$ $1$ $2$ $1$
$N$ $2$ $1$ $1.25$
She makes a profit of $Rs. 600$ and $Rs. 400$ on items $M$ and $N$ respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?

Let

$x$ and

$y$ be the number of items

$M$ and

$N$ respectively.

Total profit on the production

$= Rs (600 x + 400 y)$
Mathematical formulation of the given problem is as follows :

Maximise

$Z = 600 x + 400 y$

subject to the constraints :

$x + 2y \leq 12$ (constraint on Machine I)

$... (1)$

$2x + y \leq 12$ (constraint on Machine II)

$... (2)$

$x + 54y \geq 5$ (constraint on Machine III)

$... (3)$

$x \geq 0, y \geq 0 ... (4)$

Let us draw the graph of constraints

$(1)$ to

$(4). ABCDE$ is the feasible region (shaded) as shown in Fig determined by the constraints

$(1)$ to

$(4)$ . Observe that the feasible region is bounded, coordinates of the corner points

$A, B, C, D$ and

$E$ are

$(5, 0) (6, 0), (4, 4), (0, 6)$ and

$(0, 4)$ respectively.

Let us evaluate

$Z = 600x + 400y$ at these corner points.

Corner point	$Z = 600x + 400y$
$(5, 0)$	$3000$
$(6, 0)$	$3600$
$(4, 4)$	$4000\leftarrow Maximum$
$(0, 6)$	$2400$
$(0, 4)$	$1600$

We see that the point

$(4, 4)$ is giving the maximum value of

$Z$ . Hence, the manufacturer has to produce

$4$ units of each item to get the maximum profit of

$Rs. 4000$ .

One kind of cake requires $200g$ of flour and $25g$ of fat, and another kind of cake requires $100g$ of flour and $50g$ of fat. Find the maximum number of cakes that can be made from $5kg$ of flour and $1\ kg$ of fat assuming that there is no shortage of the other ingredients used in making the cakes.

Let number of first kind of cake is

$X$ and another kind of cake is

$Y$ .

So, total flour required

$=200X+100Y\ g$

and total fat required

$=25X+50Y\ g$

Since, maximum flour available is

$5kg=5000g$

$\therefore 200X+100Y\leq 5000$

$\Rightarrow 2X+Y\leq 50\ \ \ ...(1)$

Also, maximum fat available is

$1kg=1000g$

$\therefore 25X+50Y\leq 1000$

$\Rightarrow X+2Y\leq 40\ \ \ . . .(2)$

Since, quantity of cakes can't be negative.

$\therefore X\geq 0, Y\geq 0\ \ \ ...(3)$

We have to maximize number of cakes that can be made.

So, Objective function is

$Z=X+Y$

After plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=X+Y$
A (0,20)	20
B (20,10)	30 (Maximum)
C (25,0)	25

So, maximum cake that can be made is

$30$ , where first kind of cake will be 20 and second kind of cake will be 10.

(Manufacturing problem) A manufacturing company makes two models $A$ and $B$ of a product. Each piece of Model $A$ requires $9$ labour hours for fabricating and $1$ labour hour for finishing. Each piece of Model $B$ requires $12$ labour hours for fabricating and $3$ labour hours for finishing. For fabricating and finishing, the maximum labour hours available are $180$ and $30$ respectively. The company makes a profit of $Rs. 8000$ on each piece of model $A$ and $Rs. 12000$ on each piece of Model $B$ . How many pieces of Model $A$ and Model $B$ should be manufactured per week to realise a maximum profit? What is the maximum profit per week?

Suppose

$x$ is the number of pieces of Model

$A$ and

$y$ is the number of pieces of Model

$B$ . Then
Total profit (in Rs)

$= 8000 x + 12000 y$
Let

$Z = 8000 x + 12000 y$
We now have the following mathematical model for the given problem.

Maximise

$Z = 8000 x + 12000 y ... (1)$

subject to the constraints :

$9x + 12y \leq180$ (Fabricating constraint)
i.e.

$3x + 4y \leq 60 ... (2)$

$x + 3y \leq 30$ (Finishing constraint)

$... (3)$

$x \geq 0, y \geq 0$ (non-negative constraint)

$... (4)$

The feasible region (shaded)

$OABC$ determined by the linear inequalities

$(2)$ to

$(4)$ is shown in Fig. Note that the feasible region is bounded.
Let us evaluate the objective function

$Z$ at each corner point as shown below:

Corner Point	$Z = 8000 x + 12000 y$
$0(0, 0)$	$0$
$A(20, 0)$	$160000$
$B(12, 6)$	$168000\leftarrow Maximum$
$C(0, 10)$	$120000$

We find that maximum value of

$Z$ is

$1,68,000$ at

$B (12, 6)$ . Hence, the company should produce

$12$ pieces of Model

$A$ and

$6$ pieces of Model

$B$ to realise maximum profit and maximum profit then will be

$Rs. 1,68,000$ .

(Allocation problem) A cooperative society of farmers has $50$ hectare of land to grow two crops $X$ and $Y$ . The profit from crops $X$ and $Y$ per hectare are estimated at $Rs. 10,500$ and $Rs. 9,000$ respectively. To control weeds, a liquid herbicide has to be used for crops $X$ and $Y$ at rates of $20$ litres and $10$ litres per hectare. Further, no more than $800$ litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?

Let

$x$ hectare of land be allocated to crop

$X$ and

$y$ hectare to crop

$Y$ . Obviously,

$x \geq 0, y \geq 0$ .

Profit per hectare on crop

$X = Rs. 10500$
Profit per hectare on crop

$Y = Rs. 9000$

Therefore, total profit

$= Rs (10500x + 9000y)$
The mathematical formulation of the problem is as follows:

Maximise

$Z = 10500 x + 9000 y$

subject to the constraints:

$x + y \leq 50$ (constraint related to land)

$... (1)$

$20x + 10y \leq 800$ (constraint related to use of herbicide)
i.e.

$2x + y \leq 80 ... (2)$

$x \geq 0, y \geq 0$ (non negative constraint)

$... (3)$

Let us draw the graph of the system of inequalities

$(1)$ to

$(3)$ . The feasible region

$OABC$ is shown (shaded) in Fig. Observe that the feasible region is bounded.

The coordinates of the corner points

$O, A, B$ and

$C$ are

$(0, 0), (40, 0), (30, 20)$ and

$(0, 50)$ respectively. Let us evaluate the objective function

$Z = 10500 x + 9000y$ at these vertices to find which one gives the maximum profit.

Corner Point	$Z = 10500x + 9000y$
$O(0, 0)$	$0$
$A(40, 0)$	$42000$
$B(30, 20)$	$495000\leftarrow Maximum$
$C(0, 50)$	$450000$

Hence, the society will get the maximum profit of

$Rs. 4,95,000$ by allocating

$30$ hectares for crop

$X$ and

$20$ hectares for crop

$Y$ .

Solve the following Linear Programming Problems graphically:
Minimise $Z = -3x + 4y$
subject to $x + 2y\leq 8, 3x + 2y \leq 12, x\geq 0, y \geq 0$ .

Objective function :

$Z=-3x+4y$

We have to minimize Z on constraints

$x+2y\leq 8$

$3x+2y\leq 12$

$x\geq 0,y\geq 0$

After plotting inequalities, we got feasible region as shown in the image.

Now, there are 3 corner points

$(0,4),(2,3)\ and (4,0).$

At (0,4), value of

$Z=-3\times 0+4\times 4=16$

At (2,3), value of

$Z=-3\times 2+4\times 3=6$

At (4,0), value of

$Z=-3\times 4+4\times 0=-12$

Hence minimum value of

$Z=-12$ .

Solve the following Linear Programming Problems graphically:
Maximise $Z = 5x + 3y$
subject to $3x + 5y\leq 15, 5x + 2y \leq 10, x \geq 0, y \geq 0$ .

solve these two equations

$3x+5y<=15$ -->1

$5x+2y<=10$ -->1

We will get

$(0,3),(2,0),(\cfrac{20}{19},\cfrac{45}{19})$

From the graph the corner points are

$(0,3)$ ,

$(\cfrac{20}{19},\cfrac{45}{19})$ and

$(2,0)$

and the corresponding values of

$Z$ are

$9$ ,

$\cfrac{235}{19}=12.36$ ,

$10$

maximum value is

$12.36$

Hence the optimal solution is

$12.36$

Solve the following Linear Programming Problems graphically :
Minimize $Z = 3x + 5y$
subject to $x + y \leq 4, x \geq 0$ and $y \ge 0$

solve these two equations

$x+y>=2$ -->1

$x+3y>=10$ -->1

We will get

$(0,2),(3,0),(\cfrac{3}{2},\cfrac{1}{2})$

From the graph the corner points are

$(0,2)$ ,

$(\cfrac{3}{2},\cfrac{1}{2})$ and

$(3,0)$

and the corresponding values of

$Z$ are

$10$ ,

$7$ ,

$9$

manimum value is

$7$

Hence the optimal solution is

$7$

Solve the following linear programming problem graphically :
Minimise $Z = 200 x + 500 y ... (1)$
subject to the constraints:
$x + 2y \geq 10 ... (2)$
$3x + 4y \leq 24 ... (3)$
$x \geq 0, y \geq 0 ... (4)$ .

The shaded region in Fig is the feasible region

$ABC$ determined by the system of constraints

$(2)$ to

$(4)$ , which is bounded. The coordinates of corner points

Corner Point	Corresponding value of $Z$
$(0, 5)$	$2500$
$(4, 3)$	$2300\leftarrow Minimum$
$(0, 6)$	$3000$

$A, B$ and

$C$ are

$(0,5), (4,3)$ and

$(0,6)$ respectively. Now we evaluate

$Z = 200x + 500y$ at these points.

Hence, minimum value of

$Z$ is

$2300$ attained at the point

$(4, 3)$ .

A manufacturer makes two types of toys $A$ and $B$ . Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
Types of Toys Machines Machines Machines
I II III
$A$ $12$ $18$ $6$
$B$ $6$ $0$ $9$
Each machine is available for a maximum of $6$ hours per day. If the profit on each toy of type $A$ is $Rs. 7.50$ and that on each toy of type $B$ is $Rs. 5$ , show that $15$ toys of type $A$ and $30$ of type $B$ should be manufactured in a day to get maximum profit.

Let's assume that number of toys of type A be X and number of toys of type B be Y.

Since, toy A need 12 minutes and toy B need 6 minutes of machine

$I$ time. Also machine

$I$ is available for maximum 6 hours ( 360 minutes ).

So,

$12X+6Y\leq 360$

$\Rightarrow 2X+Y\leq 60\ \ \ ...(1)$

Since, toy A need 18 minutes and toy B need 0 minutes of machine

$II$ time. Also machine

$I$ is available for maximum 6 hours ( 360 minutes ).

So,

$18X+0Y\leq 360$

$\Rightarrow X\leq 20\ \ \ ...(2)$

Since, toy A need 6 minutes and toy B need 9 minutes of machine

$III$ time. Also machine

$III$ is available for maximum 6 hours ( 360 minutes ).

So,

$6X+9Y\leq 360$

$\Rightarrow 2X+3Y\leq 120\ \ \ ...(3)$

Since, count of toys can't be negative.

$\therefore X\geq 0\ and\ Y\geq 0\ \ \ ...(4)$

Now, profit on toy A is

$7.50\ Rs$ and profit on toy B is

$5\ Rs$

So, total profit

$(Z)=7.5X+5Y$

We have to maximize the total profit of manufacturer.

After plotting all the constraints give by equation (1), (2), (3) and (4), we get the feasible region as shown in the image.

Corner points	Value of $Z=7.5X+5Y$
A (0, 40)	200
B (15, 30)	262.50 (Maximum)
C (20, 20)	250
D (20, 0)	150

Hence maximum profit that manufacturer can earn is

$262.50\ Rs$ when 15 toys of type A and 30 toys of type B manufactured.

Two godowns $A$ and $B$ have grain capacity of $100$ quintals and $50$ quintals respectively. They supply to $3$ ration shops, $D, E$ and $F$ whose requirements are $60, 50$ and $40$ quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs)
From/ To $A$ $B$
$D$ $6$ $4$
$E$ $3$ $2$
$F$ $2.50$ $3$
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

Let's assume that godown A supplies

$X$ quintals grain to ration shop D and

$Y$ quintals grain to ration shop E.

So, godown A will supply remaining grains

$100-X-Y$ quintals to ration shop F.

Also, godown B will supply

$60-X$ quintals grain to ration shop D,

$50-Y$ quintals grain to ration shop E and

$40-(100-X-Y)=X+Y-60$ quintals grain to ration shop F.

Now total transportation cost is

$Z=6\times X+3\times Y+2.5\times (100-X-Y)+4\times (60-X)+2\times (50-Y)+3\times (X+Y-60)$

$\Rightarrow Z=6X+3Y+250-2.5X-2.5Y+240-4X+100-2Y+3X+3Y-180$

$\Rightarrow Z=2.5X+1.5Y+410\ \ \ ...(1)$

Now, Since godown A can supply maximum 60 quintals to ration shop D and 50 quintals to ration shop E and have maximum 100 quintals capacity to supply.

$\therefore X+Y\leq 100, X\leq 60\ and\ Y\leq 50\ \ \ ...(2)$

Also, if godown A supplies all 40 quintals to ration shop F, then remaining 60 quintals will be supplied to ration shop D and E.

So,

$X+Y\geq 60\ \ \ ...(3)$

Since, X and Y is amount of grains. It can never be negative.

So,

$X\geq 0, Y\geq 0\ \ \ ...(4)$

We have to minimise transportation cost given by equation (1).

After plotting all the constraints given by equation (2), (3) and (4), we got the feasible region as shown in the image.

Corner points	Value of $Z=2.5X+1.5Y+410$
A (60, 0)	560
B (60, 40)	620
C (50, 50)	610
D (10, 50)	510 (Minimum)

Hence minimum transportation cost is

$510\ Rs$ . Transporter needs to supply as follows:

Godown A to Ration shop D

$=X=10$ quintals

Godown A to Ration shop E

$=Y=50$ quintals

Godown A to Ration shop F

$=100-X-Y=40$ quintals

Godown B to Ration shop D

$=60-X=50$ quintals

Godown B to Ration shop E

$=50-Y=0$ quintals

Godown B to Ration shop F

$=X+Y-60=0$ quintals

A manufacturer produces nuts and bolts. It takes $1$ hour of work on machine $A$ and $3$ hours on machine $B$ to produce a package of nuts. It takes $3$ hours on machine $A$ and $1$ hour on machine $B$ to produce a package of bolts. He earns a profit of $Rs. 17.50$ per package on nuts and $Rs. 7.00$ per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most $12$ hours a day?

Let number of package of nuts to be made is

$X$ and number of package of bolts to be made is

$Y$

Since, nuts package requires

$1$ hours and bolts package requires

$3$ hours of machine A work time. Also, there is maximum

$12$ hours of machine A time available.

$\therefore X+3Y\leq 12\ \ \ ...(1)$

Since, nuts package requires

$3$ hours and bolts package requires

$1$ hours of machine B work time. Also, there is maximum

$12$ hours of machine B time available.

$\therefore 3X+Y\leq 12\ \ \ ...(1)$

Since, count of package can't be negative.

$\therefore X\geq 0,Y\geq 0\ \ \ ...(3)$

We have to maximize profit of the manufacturer.

Here, profit on nuts package is

$17.5\ Rs$ and on bolts package is

$7\ Rs$

So, objective function is

$Z=17.5X+7Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=17.5X+7Y$
A $(0,4)$	$28$
B $(3,3)$	$73.5$ (maximum)
C $(4,0)$	$70$

Hence, manufacturer should produce 3 packages of each to maximize his profit.

A dietician wishes to mix together two kinds of food $X$ and $Y$ in such a way that the mixture contains at least $10$ units of vitamin A, $12$ units of vitamin B and $8$ units of vitamin C. The vitamin contents of one kg food is given below:
Food Vitamin A Vitamin B Vitamin C
$X$ $1$ $2$ $3$
$Y$ $2$ $2$ $1$
One kg of food $X$ costs $Rs. 16$ and one kg of food $Y$ costs $Rs. 20$ . Find the least cost of the mixture which will produce the required diet?

Let's assume that mixture contains

$X\ kg$ of food X and

$Y\ kg$ of food Y.

Cost of 1 kg of food X

$=16\ Rs$

Cost of 1 kg of food Y

$=20\ Rs$

So, Total cost (C)

$=16X+20Y\ Rs\ \ \ \ ...(1)$

Now, food X contains

$1\ units$ and food Y contains

$2\ units$ of Vitamin A.

So, Total Vitamin A

$=X+2Y\ units$

Since, minimum requirement of Vitamin A is 10 units.

$\therefore X+2Y\geq 10\ \ \ \ ...(2)$

Again, food X contains

$2\ units$ and food Y contains

$2\ units$ of Vitamin B.

So, Total Vitamin B

$=2X+2Y\ units$

Since, minimum requirement of Vitamin B is 12 units.

$\therefore 2X+2Y\geq 12$

$\Rightarrow X+Y\geq 6\ \ \ \ ...(3)$

Also, food X contains

$3\ units$ and food Y contains

$1\ units$ of Vitamin C.

So, Total Vitamin C

$=3X+Y\ units$

Since, minimum requirement of Vitamin C is 8 units.

$\therefore 3X+Y\geq 8\ \ \ \ ...(4)$

Since, amount of food can never be negative.

So,

$X\geq 0, Y\geq 0\ \ \ \ ...(5)$

We have to minimise the cost of mixture given in equation (1) subject to the constraints given in (2),(3), (4) and (5).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $Z=16X+20Y$
A (0,12)	160
B (3,6)	116
C (9,2)	112 (minimum)
D (10,0)	160

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, plot the region

$Z< 112$ to check if there exist some points in feasible region where value can be less than 112.

$\Rightarrow 16X+20Y<112$

$\Rightarrow 4X+5Y<28$ .

Since there is no common points between the feasible region and the region which contains

$Z<112$ (See the image). So

$112\ Rs$ is the minimum cost.

There are two types of fertilisers $F_{1}$ and $F_{2}\cdot F_{1}$ consists of $10$ % nitrogen and $6$ %phosphoric acid and $F_{2}$ consists of $5$ % nitrogen and $10$ % phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast $14\ kg$ of nitrogen and $14\ kg$ of phosphoric acid for her crop. If $F_{1}$ costs $Rs. 6/kg$ and $F_{2}$ costs $Rs. 5/kg$ , determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Let's assume that amount of fertiliser

$F_1$ used be

$X$ kg and amount of fertiliser

$F_2$ used be

$Y$ kg.

Since,

$F_1$ consist

$10\%$ and

$F_2$ consist

$5\%$ nitrogen. Also, farmer needs at least 14 kg of nitrogen.

$\therefore (10\ \%\ of\ X)+(5\ \%\ of\ Y)\geq 14$

$\Rightarrow \dfrac{10X}{100}+\dfrac{5Y}{100}\geq 14$

$\Rightarrow 2X+Y\geq 280\ \ \ ...(1)$

Again since,

$F_1$ consist

$6\%$ and

$F_2$ consist

$10\%$ phosphoric acid. Also, farmer needs at least 14 kg of phosphoric acid.

$\therefore (6\ \%\ of\ X)+(10\ \%\ of\ Y)\geq 14$

$\Rightarrow \dfrac{6X}{100}+\dfrac{10Y}{100}\geq 14$

$\Rightarrow 3X+5Y\geq 700\ \ \ ...(2)$

Since amount of fertilisers can't be negative.

$\therefore X\geq 0, Y\geq 0\ \ \ ...(3)$

Since,

$F_1$ costs

$6$ Rs per kg and

$F_2$ costs

$5$ Rs per kg.

So, total costs

$(Z)=6X+5Y$

We have to minimise total costs of farmer subjects to the constraints given by (1), (2) and (3).

After plotting all the constraints given by equation (1), (2) and (3) we got the feasible region as shown in the image.

Corner points	Value of $Z=6X+5Y$
A $(0,180)$	$1400$
B $(100,80)$	$1000$ (minimum)
C $(\dfrac{700}{3},0)$	$1400$

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, after plotting the region

$Z< 1000$ to check if there exist some points in feasible region where value can be less than 1000.

$\Rightarrow 6X+5Y<1000$

Since there is no other points than B are common points between the feasible region and the region which contains

$Z<1000$ . So 1000 is the minimum cost.

Hence, for minimum costs,

$F_1$ should be

$100$ kg and

$F_2$ should be

$80$ kg. In this case, minimum costs to farmer will be

$1000\ Rs$

Use the graphical method to solve each of the following LP problems.
LESCO Engineering produces chairs and tables. Each table takes four hours of labour from the carpentry department and two hours of labour from the finishing department. Each chair requires three hours of carpentry and one hour of finishing.
During the current week, $240$ hours of carpentry time are available and $100$ hours of finishing time. Each table produced gives a profit of $E70$ and each chair a profit of $E50$ . How many chairs and tables should be made in order to maximize profit?

Department	Tables	Chairs	Available Hours
Carpentry	4	3	240
Painting	2	1	100
Profit	7	5

Let

${ x }_{ 1 }$ =number of tables to be produced

${ x }_{ 2 }$ =number of chairs to be produced

Maximize

$P=7{ x }_{ 1 }+5{ x }_{ 2 }$ (objective Function)

${ 4x }_{ 1 }+3{ x }_{ 2 }\le 240 \\ 2{ x }_{ 1 }+{ x }_{ 2 }\le 100\\ { x }_{ 1 }\ge 0, { x }_{ 2 }\ge 0$

To find the optional solution to this LP using a graphical method we first identify the region of feasible solutions

The corner point are (0,0) ,(50,0) ,(30,40) and (0,80)

Testing these corner points on

$7{ x}_{1 }+5{ x}_{2 }$

Corner Point	Profit $(P={ 7 }{ x_1 }+{ 5 }{ x_2 })$
(0,0)	0
(50,0)	350
(30,40)	410 $\rightarrow$ Maximum
(0,80)	400

At the point (30,40) produces the highest profit i.e Rs.410

$(x,y)=(30,40)$

$P=410Rs$

If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

Let's assume that bags of brand P needed be

$X$ and brand Q be

$Y$

Since bag of brand P contains 1 kg and bag of brand Q contains 2 kg of Phosphoric acid. Also, garden needs at least 240 kg of phosphoric acid.

$\therefore X+2Y\geq 240\ \ \ ...(1)$

Since bag of brand P contains 3 kg and bag of brand Q contains 1.5 kg of potash. Also, garden needs at least 270 kg of potash.

$\therefore 3X+1.5Y\geq 270$

$\Rightarrow 2X+Y\geq 180\ \ \ ...(2)$

Since bag of brand P contains 1.5 kg and bag of brand Q contains 2 kg of chlorine. Also, garden needs at most 310 kg of chlorine.

$\therefore 1.5X+2Y\leq 310\ \ \ ...(3)$

Since count of bags can't be negative.

$\therefore X\geq 0, Y\geq 0\ \ \ ...(4)$

Now, Bag of brand P contains 3 kg of Nitrogen and Bag of brand Q contains 3.5 kg of Nitrogen.

So, total Nitrogen contents

$Z=3X+3.5Y$

We have to maximise the amount of Nitrogen added to the garden.

After plotting all the constraints given by equation (1), (2), (3) and (4) we get the feasible region as shown in the image.

Corner points	Value of $Z=3X+3.5Y$
A (20, 140)	550
B (40, 100)	470
C (140, 50)	595 (maximum)

Hence, to maximise the amount of nitrogen added to the garden, grower should use

$140$ bags of brand P and

$50$ bags of brand Q. In this case maximum amount of nitrogen added to the garden will be

$595\ Kg$

A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes $2$ hours on grinding/cutting machine and $3$ hours on the sprayer to manufacture a pedestal lamp. It takes $1$ hour on the grinding/cutting machine and $2$ hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most $20$ hours and the grinding/cutting machine for at the most $12$ hours. The profit from the sale of a lamp is $Rs. 5$ and that from a shade is $Rs. 3$ . Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?

Let number of pedestal lamps to be made is

$X$ and number of wooden shades to be made is

$Y$

Since, pedestal lamps requires

$2$ hours and wooden shades requires

$1$ hours of grinding/cutting time. Also, there is maximum

$12$ hours of grinding/cutting time.

$\therefore 2X+Y\leq 12\ \ \ ...(1)$

Since, pedestal lamps requires

$3$ hours and wooden shades requires

$2$ hours for spayer. Also, there is maximum

$20$ hours for spayer.

$\therefore 3X+2Y\leq 20\ \ \ ...(2)$

Since, count of objects can't be negative.

$\therefore X\geq 0,Y\geq 0\ \ \ ...(3)$

We have to maximize profit of the industry.

Here, profit on pedestal lamps is

$5\ Rs$ and on wooden shades is

$3\ Rs$

So, objective function is

$Z=5X+3Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=5X+3Y$
A (0,10)	30
B (4,4)	32 (maximum)
C (6,0)	30

Hence, industry should produce 4 pedestal lamps and 4 wooden shades in a day to maximise his profit. Also, maximum profit will be

$32\ Rs$

An oil company has two depots $A$ and $B$ with capacities of $7000\ L$ and $4000\ L$ respectively. The company is to supply oil to three petrol pumps, $D, E$ and $F$ whose requirements are $4500L, 3000L$ and $3500L$ respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Distance in (km.)
From/ To $A$ $B$
$D$ $7$ $3$
$E$ $6$ $4$
$F$ $3$ $2$
Assuming that the transportation cost of $10$ litres of oil is $Re. 1$ per km, how should the delivery be scheduled in order that the transportation cost is minimum?
What is the minimum cost?

Let's assume that depot A supplies

$X$ L oil to petrol pump D and

$Y$ L oil to petrol pump E.

So, depot A will supply remaining oil

$7000-X-Y$ L to petrol pump F.

Also, depot B will supply

$4500-X$ L oil to petrol pump D,

$3000-Y$ L oil to petrol pump E and

$3500-(7000-X-Y)=X+Y-350$ L oil to petrol pump F.

Since cost of transportation of 10 L oils for 1 km is 1 Rs.

So, per L oils for 1 km, cost =

$\dfrac{1}{10}=0.1\ Rs$

Now total transportation cost is

$Z=0.7\times X+0.6\times Y+0.3\times (7000-X-Y)+0.3\times (4500-X)+0.4\times (3000-Y)+0.2\times (X+Y-3500)$

$\Rightarrow Z=0.7X+0.6Y+2100-0.3X-0.3Y+1350-0.3X+1200-0.4Y+0.2X+0.2Y-700$

$\Rightarrow Z=0.3X+0.1Y+3950\ \ \ ...(1)$

Now, Since depot A can supply maximum 4500 L to petrol pump D and 3000 L to petrol pump E and have maximum 7000 L capacity to supply.

$\therefore X+Y\leq 7000, X\leq 4500\ and\ Y\leq 3000\ \ \ ...(2)$

Also, if depot A supplies all 3500 L to petrol pump F, then remaining 3500 L will be supplied to petrol pump D and E.

So,

$X+Y\geq 3500\ \ \ ...(3)$

Since, X and Y is amount of oil. It can never be negative.

So,

$X\geq 0, Y\geq 0\ \ \ ...(4)$

We have to minimise transportation cost given by equation (1).

After plotting all the constraints given by equation (2), (3) and (4), we got the feasible region as shown in the image.

Corner points	Value of $Z=0.3X+0.1Y+3950$
A (500, 3000)	4400 (Minimum)
B (3500, 0)	5000
C (4500, 0)	5300
D (4500, 2500)	5550
E (4000, 3000)	5450

Hence minimum transportation cost is

$4400\ Rs$ . Transporter needs to supply as follows:

Depot A to petrol pump D

$=X=500$ L

Depot A to petrol pump E

$=Y=3000$ L

Depot A to petrol pump F

$=7000-X-Y=3500$ L

Depot B to petrol pump D

$=4500-X=4000$ L

Depot B to petrol pump E

$=3000-Y=0$ L

Depot B to petrol pump F

$=X+Y-3500=0$ L

A fruit grower can use two types of fertilizer in his garden, brand $P$ and brand $Q$ .The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least $240\ kg$ of phosphoric acid, at least $270\ kg$ of potash and at most $310\ kg$ of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
Kg per bag
Brand $P$ Brand $Q$
Nitrogen $3$ $3.5$
Phosphoric acid $1$ $2$
Potash $3$ $1.5$
Chlorine $1.5$ $2$

Let's assume that bags of brand P needed be

$X$ and brand Q be

$Y$

Since bag of brand P contains 1 kg and bag of brand Q contains 2 kg of Phosphoric acid. Also, garden needs at least 240 kg of phosphoric acid.

$\therefore X+2Y\geq 240\ \ \ ...(1)$

Since bag of brand P contains 3 kg and bag of brand Q contains 1.5 kg of potash. Also, garden needs at least 270 kg of potash.

$\therefore 3X+1.5Y\geq 270$

$\Rightarrow 2X+Y\geq 180\ \ \ ...(2)$

Since bag of brand P contains 1.5 kg and bag of brand Q contains 2 kg of chlorine. Also, garden needs at most 310 kg of chlorine.

$\therefore 1.5X+2Y\leq 310\ \ \ ...(3)$

Since count of bags can't be negative.

$\therefore X\geq 0, Y\geq 0\ \ \ ...(4)$

Now, Bag of brand P contains 3 kg of Nitrogen and Bag of brand Q contains 3.5 kg of Nitrogen.

So, total Nitrogen contents

$Z=3X+3.5Y$

We have to minimise the amount of Nitrogen added to the garden.

After plotting all the constraints given by equation (1), (2), (3) and (4) we get the feasible region as shown in the image.

Corner points	Value of $Z=3X+3.5Y$
A (20, 140)	550
B (40, 100)	470 (Minimum)
C (140, 50)	595

Hence, to minimise the amount of nitrogen added to the garden, grower should use

$40$ bags of brand P and

$100$ bags of brand Q. In this case minimum amount of nitrogen added to the garden will be

$470\ Kg$

Holiday Mean Turkey Ranch is considering buying two different types of turkey feed. Each feed contains, in varying proportions, some or all of the three nutritional ingredients essential for fattening turkeys. Brand Y feed costs the ranch $$0.2$ per pound. Brand Z costs $$.03$ per pound. The rancher would like to determine the lowest-cost diet that meets the minimum monthly intake requirement for each nutritional ingredient.
The following table contains relevant information about the composition of brand Y and brand Z feeds, as well as the minimum monthly requirement for each nutritional ingredient per turkey.
Composition of Each Pound of Feed
Ingredient Brand Y Feed Brand Z Feed Minimum Monthly Requirement
A $5$ oz $10$ oz $90$ oz
B $4$ oz $3$ oz $48$ oz
C $.5$ oz $0$ $1.5$ oz
Cost/lb $$0.2$ $$.03$

If we let

${X}_{1}=$ number of pounds of brand Y feed purchased

${X}_{2}=$ number of pounds of brand Z feed purchased
then we may proceed to formulate this linear programming problem as follows:
Minimize cost (in cents)

$=2{X}_{1}+3{X}_{2}$
subject to these constraints

$5{X}_{1}+10{X}_{2}\ge 90$ oz (ingredient A constraint)

$4{X}_{1}+3{X}_{2}\ge 48$ oz (ingredient B constraint)

$\cfrac{1}{2}{X}_{1}\ge 1\cfrac{1}{2}$ oz (ingredient C constraint)
Figure illustrates these constraints
The iso-cost line approach may be used to solve LP minimization problems such as that of the Holiday Meal Turkey Ranch. As with iso-profit lines, we need not compute the cost lines. The lowest cost line (that is, the one closest in toward the origin) to couch the feasible region provides us with the optimal solution corner.
For example, we start in figure by drawing a

$54$ cents cos line, namely,

$54=2{X}_{1}+3{}_{2}$ . Obviously, there are many points in the feasible region that would yield a lower total cost. We proceed to move our iso-cost line toward the lower left, in plane parallel to the

$54$ cent solution line. The lase point we touch while still in contact with the feasible region is the same as corner point b of figure. It has the coordinates (

${X}_{1}=8.4,{X}_{2}=4.8$ ) and an associated cost of

$31.2$ cents

A farmer plans to mix two types of food to make a mix of low cost feed for the animals in his farm. A bag of food $A$ costs $$10$ and contains $40$ units of proteins, $20$ units of minerals and $10$ units of vitamins. A bag of food $B$ costs $$12$ contains $30$ units of proteins, $20$ units of minerals and $30$ units of vitamins. How many bags of food $A$ and $B$ should the consumed by the animals each day in order to meet the minimum daily requirements of $150$ units of proteins, $90$ units of minerals and $60$ units of vitamins at a minimum cost?

Let

$x$ be the number of bags of food

$A$ and

$y$ the number of bags of food

$B$
Cost

$C(x,y)=10x+12y$
The solution set of the system of inequalities above and the vertices of the feasible solution set obtained are shown below:
Vertices:

$A$ at intersection (x-axis) coordinates of

$A:(6,0)$

$B$ at intersection of

$10x+30y=60$ coordinates of

$B:(15/4,4/3)$

$C$ at intersection of

$40x+30y=150$ and

$20x+20y=90$ coordinates of

$D:(0,5)$
Evaluate the cost

$c(x,y)=10x+12y$ at each one of the vertices

$A(x,y),B(x,y),C(x,y)$ and

$D(x,y)$
At

$A(6,0):c(6,0)=10(6)+12(0)=60$
At

$B(15/4,3/4):c(15/4,3/4)=10(15/4)+12(3/4)=46.5$
At

$C(3/2,3):c(2,3)=10(3/2)+12(3)=51$
At

$D(0,5):c(0,5)=10(0)+12(5)=60$
The cost

$c(x,y)$ is minimum at the vertex

$B(15/4,3/4)$ where

$x=15/4=3.75$ and

$y=3/4=0.75$
Hence

$3.75$ bags of food

$A$ and

$0.75$ bags of food

$B$ are needed to satisfy the minimum daily requirements in terms of proteins, minerals and vitamins at the lowest possible cost.
1032489_849321_ans_538ec92c724648a9b695e395eea39fd2.PNG

1032489_849321_ans_538ec92c724648a9b695e395eea39fd2.PNG

A store sells two types of toys, $A$ and $B$ . The store owner pays $$8$ and $$14$ for each one unit of toy $A$ and $B$ respectively. One unit of toys $A$ yields a profit of $$2$ while a unit of toys $B$ yields a profit of $$2$ . The store owner estimates that no more than $2000$ toys will be sold every month and he does not plan to invest more than $$20,000$ in inventory of these toys. How many units each type of toys should be stocked in order to maximize his monthly total profit?

Let

$x$ be the total number of toys

$A$ and

$y$ the number of toys

$B$ ;

$x$ and

$y$ cannot be negative, hence

$x\ge 0$ and

$y\ge 0$
The store owner estimates that no more than

$2000$ toys will be old every month.

$x+y\le 2000$
One unit of toys

$A$ yields a profit of

$$2$ while a unit of toys

$B$ yields a profit of

$$3$ , hence the total profit

$P$ is given by

$P=2x+3y$
The store owner pays

$$8$ and

$$14$ for each one unit of toy

$A$ and

$B$ respectively and he does not plant to invest more than

$$20,000$ in inventory of these toys

$8x+14y\le 20,00$
What do we have to solve?
Find

$x$ and

$y$ so that

$P=2x+3y$ is maximum under the conditions

$\begin{cases} x\ge 0 \\ x\ge 0 \\ x+y\le 2000 \\ 8x+14y\le 20,000 \end{cases}$
The solution set of the system of inequalities given above and the vertices of the region obtained are shown below:
Vertices of the solution set

$A$ at

$(0,0)$

$B$ at

$(0.1429)$

$C$ at

$(1333,667)$

$D$ at

$(2000,0)$
Calculate the total profit

$P$ at each vertex

$P(A)=2(0)+3(0)=0$

$P(B)=2(0)+3(1429)=4287$

$P(C)=2(1333)+3(667)=4667$

$P(D)=2(2000)+3(0)=4000$
The maximum profit is at vertex

$C$ with

$x=1333$ and

$y=667$
Hence the store owner has to have

$1333$ toys of type

$A$ and

$667$ toys of type

$B$ in order to maximize his profit
1032459_849210_ans_dee16aa4532c46abbd72245a7cec74a9.PNG

Solve using graphical method:
Maximize: $Z=4x+5y$
Subject to:
$2x+5y\le 25$
$6x+5y\le 45$
$49$
$x\ge 0, y\ge 0$

To solve the above linear programming model using the graphical method, we shall turn each constraints inequality to equation and set each variable equal to zero (0) to obtain two (2) coordinate points for each equation (i.e using double intercept form). Having obtained all the coordinate points, we shall determine the range of our variables which enables us to know the appropriate scale to use for our graph. Thereafter, we shall draw the graph and join all the coordinate points with required straight line.

$2x+5y=25$ (Constraint 1)
When

$x=0, y=5$ and when

$y=0,x=12.5$

$6x+5y=45$ (Constraint 2)
When

$x=0, y=9$ and
when

$y=0, x=7.5$
Minimum value of

$x$ is

$x=0$
Maximum value of

$x$ is

$x=12.5$
Range of

$x$ is

$0\le x\le 12.5$
Minimum value of

$y$ is

$y=0$
Maximum value of

$y$ is

$y=9$
The constrains give a set of feasible solutions as graphed. To solve the linear programming problem, we must now find the feasible solution that makes the objective function as large as possible. Some possible solution are listed below

Feasible solution (A point in the solution set of the system)	Objective function $Z=4x+5y$
$(2,3)$	$4(2)+5(3)=8+15=23$
$(4,2)$	$4(4)+5(2)=16+10=26$
$(5,1)$	$4(5)+5(1)=20+5=25$
$(7,0)$	$4(7)+5(0)=28+0=28$
$(0,5)$	$4(0)+5(5)=0+25=25$

In this list, the point that makes the objective function the largest is

$(7,0)$ . But, is this the largest for all feasible solutions? How about

$(6,1)$ ? or

$(5,3)$ ? IT turns out that

$(5,3)$ provide the maximum value;

$4(5)+5(3)=20+15=35$
Hence, the maximum profit at point

$(5,3)$ and it is the objective functions which have optimal values
1032439_848862_ans_479f4b8987a7411db9c34c57d7923b6d.PNG

Solve the following problem:
Maximize $Z=-P{x}_{1}+4{x}_{2}$
Subject to: $-3{x}_{1}+{x}_{2}\le 6$
${x}_{1}+2{x}_{2}\le 4$
${x}_{2}\le -3$
No lower bound constraint for ${x}_{1}$

The third constraint can be re-written as

$-{x}_{2}\ge 3$ . The solution space satisfying constraints is shown shaded in the figure.

$A(-3,-3)$ and

$B(10,-3)$ value of the objective functions

$Z=-{x}_{1}+4{x}_{2}$ at these vertices are

$Z(A)=3-12=-9$ ,

$Z(B)=-10-12=-22$
Thus the maximum value of

$Z$ occurs at

$A$ . Hence the solution to the problem is

${x}_{1}=-2,{x}_{2}-3;{Z}_{max}=-9$

Suppose a manufacturer of printed circuits has a stock of $200$ resistors, $120$ transistors and $150$ capacitors and is required to produce two types of circuits.
Type A requires $20$ resistors, $10$ transistors and $10$ capacitors
Type B requires $10$ resistors, $20$ transistors and $30$ capacitors.
If the profit on type $A$ circuits is $E5$ and that on type $B$ circuits is $E12$ , how many of each circuit should be produced in order to maximize profit?

Given:

	Resistors	Transistors	Capacitors
Type A $(x)$	20	10	10
Type B $(y)$	10	20	30
Total	200	120	150

Maximize

$Z=5x+12y$

Solution:

Let type A be

$x$ and let type be

$y$

$i)20x+10y\le 200\\ ii)10x+20y\le 120\\ iii)10x+30y\le 150$ constraints

$20x+10y=200$

$x$	0	10
$y$	20	0

Point A(0,20), B(10,0)
ii)

$10x+20y=120$

$x$	0	12
$y$	6	0

Point C(0,6) D(12,0)
iii)

$10x+30y=150$

$x$	0	15
$y$	5	0

Point E(0,5) F(15,0)

iv)

$\dfrac { \dfrac { 20x+10y=200\\ 10x+20y=120 }{ 400+200=4000\\ \pm 100x\pm 200=\pm 1200 } }{ 300=2800\\ x=9.3 } \\ 20x+10y=200\\ 20(9.3)+10y=200\\ \quad \quad \quad 10y=14\\ \quad \quad \quad \quad y=1.4$ G(9.3,1.4)

$\dfrac { 10x+20y=120\\ \pm 10x\pm 30y=\pm 150 }{ -10y=-30\\ \quad \quad \quad y=3\\ } \\ 10x+20y=120\\ 10x+60=120\\ 10x=60\\ \quad x=6$ Point H(6,3)

The feasible region

$CGBO$

Points	$z=5x+12y$
O(0,0)	0
C(0,6)	72 $\rightarrow$ Maximum
G(9.3,1.4)	63.3
B(10,0)	50

The maximum value is 72 and the points

$(x,y)$ are

(0,6)

$\Rightarrow$

$x=$

$y=6$

A company produces two types of tables, ${T}_{1}$ and ${T}_{2}$ . It takes $2$ hours to produce the parts of one unit of ${T}_{1}$ , $1$ hour to assemble and $2$ hours ot polish. It takes $4$ hours to produce the parts of one unit of ${T}_{2}$ , $2.5$ hour to assemble and $1.5$ hours of polish. Per month, $7000$ hours are available for producing the parts, $4000$ hours for assembling the parts and $5500$ hours for polishing the tables. The profit per unit of ${T}_{1}$ is $$90$ and per unit of ${T}_{2}$ is $$110$ . How many of each type of tables should be produced in order to maximize the total monthly profit?

Let

$x$ be the number of tables of type

${T}_{1}$ and

$y$ the number of tables of type

${T}_{2}$ . Profit

$P(x,y)=90x+110y$

$\begin{cases} x\ge 0 \\ x\ge 0 \\ 2x+4y\le 7000 \\ x+2.5y\le 4000 \\ 2x+1,5y\le 5500 \end{cases}$
The solution set of the system of inequalities above the vertices of the feasible solution set obtained are shown below.

$A$ at

$(0,0)$

$B$ at

$(0,1600)$

$C$ at

$(1500,1000)$

$D$ at

$(2300,600)$

$E$ at

$(2750,0)$
Evaluate profit

$P(x,y)$ at each vertex

$A$ at

$(0,0):P(0,0)=0$

$B$ at

$(0.1600):P(0,1600)=90(0)+110(1600)=176000$

$C$ at

$(1500,1000):P(1500,1000)=90(2500)+110(1000)=245000$

$D$ at

$(2300,600):P(2300,600)=90(2300)+110(600)=273000$

$E$ at

$(2750,0):P(2750,0)=90(2750)+110(0)=247500$
The maximum profit of

$$273000$ is at vertex

$D$ . Hence the company needs to product

$2300$ tables of type

${T}_{1}$ and

$600$ tables of type

${T}_{2}$ in order to maximize its profit.
1032468_849253_ans_b380c500d9ce41408b9b2512087e99ce.PNG

Use the graphical method to solve each of the following LP problems.
A factory manufactures two products, each requiring the use of three machines. The first machine can be used at most $70$ hours; the second machine at most $40$ hours; and the third machine at most $90$ hours. The first product requires $2$ hours on Machine 1, $1$ hour on Machine 2, and $1$ hour on Machine 3; the second product requires $1$ hour each on machine 1 and 2 and $3$ hours on MachineIf the profit in $E40$ per unit for the first product and $E60$ per unit for the second product, how many units of each product should be manufactured to maximize profit?

Given

Machine	Product A	Product B	Time
1	2	1	70
2	1	1	40
3	1	3	90
Profit	40	60

The constraints for given data

For Machine:1

$2x+y\le$ 70

$x+y\le$ 40

$x+3y\le$ 90

$x\ge$ 0 ,

$y\ge$ 0

To maximize:Total profit(

$z$ )=

$40x+60y$

The bounded area is OEGB

For

$z=40x+60y$

Point $(x,y)$	z value $(40x+60y)$
O(0,0)	0
E(0,30)	1800
G(20,30)	2600 $\rightarrow$ maximum
B(0,35)	2100

So at value G(20,30) the profit is maximum and the maximum value is 2600

Using graphical method
Find the corner points for
$2x+5y\le 25$
$6x+5y\le 45$
$x\ge 0,y\ge 0$

Some corner points can usually found by inspection. In this case, we can see

$A=(0,0)$ and

$D(0,5)$ . Some corner points may require some work with boundary lines (uses equations of boundaries not the inequalities giving the regions).
Point C
System:

$2x+5y=25...(1)$

$6x+5y=45.....(2)$

$(1)-(2)$

$R-4x=-20$

$\Rightarrow$

$x=5$
If

$x=5$ , then from (1) or (2)

$y=3$
Point B:
System:

$t=0.....(1)$

$6x+5y=45....(2)$
Solve by substitution:

$\Rightarrow$

$6x+5(0)=45=7.5$
The corner points for example

$7$ are

$(0,0),(0,5),(7.5,0)$ and

$5,3)$
Convex sets and corner points lead us to a method for solving certain linear programming problems.
1028274_848870_ans_08a9a796081440099e76b1e49215e5e7.PNG

1028274_848870_ans_08a9a796081440099e76b1e49215e5e7.PNG

Use graphical method
Minimize: $z=600x+900y$
Subject to:
$40x+60y\ge 480$
$30x+15y\ge 180$
$x\ge 0,y\ge 0$

If we let

$z=Rs.8100$ , then:

$8100=600x+900y,(x=0,y=9)(x=13.5,y=0)$
The resultant trial cost is shown in the figure.
The line is parallel to the boundary line BC. The lowest acceptable cost solution will be coincidental with the line BC making point B, point C and any other points on the line BC optical. Multiple optimum solutions present the manager with choice and hence some flexibility
1161463_849068_ans_a16d3c4e1c9e4b9b955683aa8da432d3.PNG

1161463_849068_ans_a16d3c4e1c9e4b9b955683aa8da432d3.PNG

Construct the graphs of the following functions.
$\displaystyle y \, = \, x \, | x | \, - \, 4x \, - \, 5.$

there are 2 cases x>0 and x<0

the respective equations are

$x^2-4x-5,-x^2-4x-5$

We get the above graph

Construct the graphs of the following functions.
$\displaystyle y \, = \, (2x \, + \, 1) \, / \, (x \, - \, 1).$

according to the question, x cannot be 1 since it is not defined therefore the domain is

$R-{0}$

when

$x$ tend to infinity then the value becomes 2

when

$x$ tends to

$1$

Hence, we get the above graph

Construct the graphs of the following functions.
$\displaystyle y \, = \, ( |1 \, - \, x \, | \, + \, 2) \, (x \, + \, 1).$

we have two cases

$x>1,x<1$

the two equations are respectively

$(x+1)^2$ ,

$2x-x^2+3$

therefore the graph is

Construct the graphs of the following functions.
$\displaystyle y \, = \, | 2x^2 \, - \, 3x \, + \, | x \, - \, 1| \, |.$

there are 2 cases x>0 and x<0

when

$x>1$ the function is

$|2x^2-2x-1|$

when

$x<1$ the function is

$|2x^2-4x+1|$

Hence, we get the above graph

Construct the graphs of the following functions.
$\displaystyle y \, = \, 1 / | x |.$

$x=0$ then the function is undefined

as it is

$|x|$ it is aways positive and symmetrical to x axis,

Hence, we get the above graph

Construct the graphs of the following functions.
$\displaystyle y \, = \, \frac{| x \, - \, 1 |}{x \, - \, 1} \, (x^2 \, + \, 3).$

$x>1$ the equation becomes

$(x^2+3)$

$x<1$ the equation becomes

$(-x^2-3)$

$x=1$ the function is not defined

Hence, we get the above graph

John has $$20,000$ to invest in three funds ${F}_{1},{F}_{2}$ and ${F}_{3}$ . Fund ${F}_{1}$ is offers a return of $2$ % and has a low risk. Fund ${F}_{2}$ offers a return of $4$ % and has a medium risk. Fund ${F}_{3}$ offers a return of $5$ % but has a high risk. To be the safe side, John invests no more than $$3000$ in ${F}_{3}$ and at least twice as much as in ${F}_{1}$ than in ${F}_{2}$ . Assuming that the rates hold till the end of the year, what amounts should he invest in each fund in order to maximize the year end return?

Let

$x$ be the amount invested in

${F}_{1}$ ,

$y$ the amount invested in

${F}_{2}$ and

$z$ the amount invested in

${F}_{1}$ .

$x+y+z=20,000$

$z=20,000-(x+y)$
Total return

$R$ of all three funds is given by

$R=2$ %

$x+4$ %

$y+5$ %

$z=0.2x+0.04y+0.05(20,000-(x+y))$
Simplifies to

$R(x,y)=1000-0.03yx-0.01y$ : This is the return to maximize
Constraints:

$x,y\ge 0$ which may be written as

$x+y\le 20,000$
John invests no more than

$$3000$ in

${F}_{3}$ , hence

$z\le 3000$
Substitute

$z$ by

$20,000-(x+y)$ in the above inequality to obtain

$20,000-(x+y)\le 3000$ which may be written as

$x+y\ge 17,000$
Let us put all the inequalities together to obtain the following system

$\begin{cases} x\ge 0 \\ x\ge 0 \\ (x+y)\le 17000 \\ x+y\le 20,000 \\ x\ge 2y \end{cases}$
The solution set of the system of inequalities above and the vertices of the feasible solution set obtained are shown.
Vertices

$A$ at intersection of

$x+y=20000$ and

$y=0$ , coordinates of

$A:(20000,0)$

$B$ at intersection of

$x+y=17000$ and

$y=0$ coordinates of

$B:(17000,0)$

$C$ at intersection of

$x+y=17000$ and

$x=2y$ coordinates of

$C:(11333,5667)$

$D$ is at intersection of

$x=2y$ and

$x+y=20000$ , coordinates of

$D:(13333,6667)$
Evaluete the return

$R(X,y)=1000-0.03x-0.01y$ at each one of the vertices

$A(x,y), B(x,y), C(x,y)$ and

$D(x,y)$
At

$A(20000,0):R(20000,0)=1000-0.03(20000)-0.01(0)=400$
At

$B(17000,0):R(17000,0)=1000-0.03(17000)-0.01(0)=490$
At

$C(11333,5667):R(11333,5667)=1000-0.03(13333)-0.01(5667)=603$
At

$D(13333,6667):R(13333,6667)=1000-0.03(13333)-0.01(6667)=533$
The return

$R$ is maximum at the vertex at

$C(11333,5667)$ where

$x=11333$ and

$y=5667$ and

$z=20,000-(x+y)=3000$
For maximum return, John has to invest

$$11333$ in fund

${F}_{1}$ ,

$$5667$ in fund

${F}_{2}$ and

$$3000$ in fund

${F}_{3}$

1032509_849428_ans_7494e7ebe0db429db4375a3375d624be.PNG

Construct the graphs of the following functions.
$\displaystyle y \, = \, x (x \, - \, 3)^2.$

the function becomes

$0$ when

$x=0$ or

$3$ , therefore, the graph is

Each month a store owner can spend at most $$100,000$ on PC's laptops. A PC costs the store owner $$1000$ and a laptop costs him $$1500$ . Each PC is sold for a profit of $$400$ while laptop is sold for a profit of $$700$ . The store owner estimates that at least $15$ PC's but no more than $80$ are sold each month. He also estimates that the number of laptops sold is at most half the PC's. How many PC's and how many laptops should be sold in order to maximize the profit?

Let

$x$ and

$y$ be the numbers of PC's and laptops respectively that should be sold.
Profit

$=400x+700y$ to maximize
Constraints

$15\le x\le 80$ least 15 PC's but no more than

$80$ are sold each month

$y\le (1/2)x$

$1000x+1500y\le 100,000$ "store owner can spend at most

$$100,000$ on PC's and laptops"

$\begin{cases} 15\le x\le 80 \\ y\ge 0 \\ y\le (1/2)x \\ 1000x+1500y\le 100,000 \end{cases}$
The solution set of the system of inequalities above and the vertices of the feasible solution set obtained are shown.
Vertices:

$A$ at intersection of

$x=15$ and

$y=0$ (x-axis) coordinates of

$A:(15,0)$

$B$ at intersection of

$x=15$ and

$y=(1/2)x$ coordinates of

$B:(15,7.5)$

$C$ at intersection of

$y=(1/2)x$ and

$1000x+1500y=100000$ coordinates of

$C:(57.14,28.57)$

$D$ at intersection of

$1000x+1500y=100000$ and

$x=80$ (y-axis) coordinates of

$D:(80,13,3)$
Evaluate the profit at each vertex

$A(15,0),P=400\times 15+700\times 0=6000$

$B(15,7.5).P=400\times 15+700\times 7.5=11250$

$C(57.14,28.57),P(400\times 57.14+700\times 28.57=42855$

$D(80,13.3),P=400\times 80+700\times 13.3=41310$
The profit is maximum for

$x=57.14$ and

$y=28.57$ but these cannot be accepted as solutions because

$x$ and

$y$ are numbers of PC's and ;laptops and must be integers. We need to select the nearest integers to

$x=57.14$ and

$y=28.57$ that are satisfy all constraints and give a maximum profit

$x=57$ and

$y=29$ do not satisfy all constraints

$x=57$ and

$y=28$ satisfy all constraints
Profit

$=500\times 57+700\times 28=42400$ , which is maximum
1032577_849539_ans_39c90799bf2b4748baedfae7f4d1ab9e.PNG

Solve the following maximal assignment problem:
Branch Manager Monthly Business
(Rs. lakh)
$A$ $B$ $C$ $D$
$P$ $11$ $11$ $9$ $9$
$Q$ $13$ $16$ $11$ $10$
$R$ $12$ $17$ $13$ $8$
$S$ $16$ $14$ $16$ $12$

Maximal assingment problem

	A	B	C	D
P	11	11	9	9
Q	13	16	11	10
R	12	17	13	8
S	16	14	16	12

To convert this problem into minimisation problem, subtract all th element from maximum value i.e. 17

Our new matris

	A	B	C	D
P	6	6	8	8
Q	4	1	6	7
R	5	0	4	9
S	1	3	1	5

Now reduce minimum value of row from all elements

	A	B	C	D
P	0	0	2	2
Q	3	0	5	6
R	5	0	4	9
S	0	3	0	5

Now subtract minimum value of column from all elements

	A	B	C	D
P	0	0	2	0
Q	3	0	5	4
R	5	0	4	7
S	0	3	0	3

Number of crossing line(bolded row and column)

$=3$

But order of matrix

$= 4$

$3 < 4$ (not optimal solution)

Now subtract uncrossed minimum element from uncrossed(unbolded element) and add at the T - point(Junction of bolded row and bolded column)

	A	B	C	D
P	0	3	2	0
Q	0	0	2	1
R	2	0	1	4
S	0	6	0	3

Number of crossing line(bolded row and column)

$=4$

But order of matrix

$= 4$

$4 = 4$ (optimal solution)

assigmnrnt of job

$P \to D$

$Q \to Q$

$R \to B$

$S \to C$

So maximum time

$9 + 13 + 17+ 12 = 51$

Mr. Anil wants to invest at most $Rs. 60,000$ in Fixed Deposit (F.D.) and Public Provident Fund (P.P.F.). He wants to invest at least $Rs. 20,000$ in F.D. and at least $Rs. 15,000$ in P.P.F. The rate of interest on F.D. is $8\% p.a.$ and that on P.P.F. is $10\% p.a.$ Formulate the above problem as L.P.P. to determine maximum yearly income.

Let the money F.D. invest in

$= x$

and money invested in

$PPF = y$

Constraint

$= x + y = 60,000$

$x \ge 20,000$

$y \ge 15,000$

Maximize

$z = 0.08x + 0.1 y$

Minimize and maximize $Z=600x+400y$
Subject to the constraints
$x+2y\le 12$
$2x+y\le 12$
$4x+5y\le 20$
$x\ge 0;y\ge 0$ by graphical method

Draw the graph of

$2x+y=12$ .....

$(i)$
We get intersection of

$(i)$ with the coordinate axes at points

$(0,12)$ and

$(6,0)$
Draw the graph of

$x+2y=12$ .....

$(ii)$

We get the intersection of

$(ii)$ with the coordinate axes at points

$(0\quad 6)$ and

$(12\quad 0)$
Draw the graph of

$4x+5y=20$ .....

$(iii)$
We get the intersection of

$(iii)$ with the coordinate axes at points

$(0,4)$ and

$(5,0)$

Common shaded region is the feasible region with corner points

$(5,0),(6,0),(4,4),(0,6),(0,4)$

Corner points	$Z=600x+400y$
$(0,4)$	$1600$ minimum
$(0.6)$	$2400$
$(4,4)$	$4000$ maximum
$(6,0)$	$3600$
$(5,0)$	$3000$

Minimum value of

$z=1600$ at

$(0,4)$
Maximum value of

$z=4000$ at

$(4,4)$

Solve the following using graphical method:
Minimize: $Z = 3x + 5y$ subject to $2x+ 3y \geq 12,$ $-x+y \leq 3$ , $x \leq 4, \,\, y \geq 3, \,\, x \geq 0, \,\, y \geq 0$

$z=3x+5y$

corners points are

$(0,0),(0,3),(4,0),\left(\dfrac{3}{2},3\right),\left(4,\dfrac{4}{3}\right)$

$z$ at

$(0,0)=0$

$z$ at

$(0,3)=0+5(3)=15$

$z$ at

$(4,0)=4(3)70=12$

$z$ at

$\left(\dfrac{3}{2},3\right)=3\left(\dfrac{3}{2}\right)+5(3)=\dfrac{39}{2}$

$z$ at

$\left(4,\dfrac{4}{3}\right)=3(4)+5\left(\dfrac{4}{3}\right)=\dfrac{56}{3}$

$\therefore$

$z$ is minimum at

$(0,0)$ and

$z=0$

1164050_979856_ans_efd589d551074b1b80771fbef804825f.png

Solve the following minimal assignment problem and hence find minimum time where $'-'$ indicates that job cannot be assigned to the machine:
Machines Processing time in hours
$A$ $B$ $C$ $D$ $E$
$M_{1}$ $9$ $11$ $15$ $10$ $11$
$M_{2}$ $12$ $9$ - $10$ $9$
$M_{3}$ - $11$ $14$ $11$ $7$
$M_{4}$ $14$ $8$ $12$ $7$ $8$

Add dummy row (

$M_5$ )

	A	B	C	D	E
$M_1$	9	11	15	10	11
$M_2$	12	9	-	10	9
$M_3$	-	11	14	11	7
$M_4$	14	8	12	7	8
$M_5$	0	0	0	0	0

Now apply row reduction

	A	B	C	D	E
$M_1$	0	2	6	1	2
$M_2$	3	0	-	1	0
$M_3$	-	4	7	4	0
$M_4$	7	1	5	0	1
$M_5$	0	0	0	0	0

Number of crosed(bold rows and columns) line

$= 4$

order of matrix

$=5$

$4 < 5$ (not optimal solution)

Now Modify matrix by subtracting minimum uncrossed(unbolded ) element from uncrossed element (unbolded ) element and adding same at T-point(element lying at junction of bolded rows and columns )

Modified element

	A	B	C	D	E
$M_1$	0	1	5	1	2
$M_2$	4	0	-	2	1
$M_3$	-	3	6	4	0
$M_4$	7	0	4	0	1
$M_5$	1	0	0	1	0

Number of crossed line(bolded rows and column)

$=5$

order of matrix

$=5$

$5 5$ (optimal solution)

assignment

$M_1 \to A$

$M_2 \to B$

$M_3 \to E$

$M_4 \to D$

$M_5 \to C$ (

$C$ cannot be ssigned to any one)

Optical solution

$= 9 + 9+7+7 = 32$

Solve the following problem graphically:
Minimise and Maximise
$z = 3x + 9y$
Subject to the constraints:
$x + 3y \leq 60$
$x + y \geq 10$
$x \leq y$
$x\geq 0, y\geq 0$ .

Let

$Z = 3x + 9y .... (1)$
Converting inequalities to equalities

$x + 3y = 60$

$x$	$0$	$60$
$y$	$20$	$0$

Points are

$(0, 20), (60, 0)$

$x + y = 10$

$x$	$0$	$10$
$y$	$10$	$0$

Points are

$(0, 10), (10, 0)$

$x - y = 0$

$x$	$0$	$10$	$20$
$y$	$0$	$10$	$20$

Points are

$(0, 0), (10, 10), (20, 20)$
Plot the graph for the set of points
The graph shows the bounded feasible region.

$ABCD$ , with corner points

$A = (10, 0), B = (5, 5), C = (15, 15)$ and

$D = (0, 20)$
To find maximum and minimum

Corner point	$Z = 3x + 9y$
$A = (0, 10)$	$90$
$B = (5, 5)$	$60$
$C = (15, 15)$	$180$
$D = (0, 20)$	$180$

From the graph maximum value of

$X$ occurs at two corner points

$C(15, 5)$ and

$D(0, 20)$ with value

$180$
and minimum occurs at point

$B(5, 5)$ with value

$60$ .

Construct the graphs of the following functions.
$\displaystyle y \, = \, (2 | x | -1) \, / \, (x \, - \, 3).$

there are two cases

$x>0$ and

$x<0$

therefore the respective functions are

$\dfrac { 2x-1 }{ x-3 }$ and

$\dfrac { 2x+1 }{ 3-x }$

when x tends to infinity the value is

$2$

when x tends to -infinity the value is

$-2$

Hence, we get the above graph

In solving LP problem : minimize $z = 6x + 10y$ , subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0,$ which constraints are reduntant?

After plotting the graph of:

$x\geq 6$

$y\geq 2$

$2x+y\geq 10$

$x,y\geq0$

We get, only one critical point of intersevtion, i.e.,

$(6,2)$

$\therefore z_{min}=6\times 6+10\times 2=56$

Minimise $Z= 3x +2y$
subject to constraints:
$x+y \geq 8$
$3x+ 5y\leq 15$
$x\geq 0, y \geq 0$

Given:

$z=3x+2y$

subject to constraints:

$x+y\ge 8\\ 3x+5y\le 15\\ x\ge 0\quad y\ge 0$ (

$1st$ quadrant)

First we will draw a graph for the constraints.

From the graph it is pretty clear that the three regions are disjoint.Hence for no value of

$'x'$ can the expression

$x=3x+2y$ be minimized .There is no domain for us to work with.Hence for this question minimization of the expression under the constraints gives is not possible.

1099636_1031306_ans_7d492f7926b14b358a7a576ee36a52e7.png

A chemical company produces two products, $X$ and $Y$ .Each unit of product $X$ requires $3$ hrs on operation $I$ and $4$ hrs on operation $II$ ,while each unit of product $Y$ requires $4$ hrs on operation $I$ and $5$ hrs on operation $II$ .
Total available time for operations $I$ and $II$ is $20$ hrs and $26$ hrs respectively.The production of each unit of product $Y$ also results in two units of a by-product $Z$ at no extra cost.
Product $X$ sells at profit of Rs. $10$ /unit, while $Y$ sells at profit of Rs. $20$ /unit.By-product $Z$ brings a unit profit of Rs. $6$ if sold;in case it cannot be sold, the destruction cost is Rs. $4$ per unit.Forecasts indicate that not more that $5$ units of $Z$ can be sold.Formulate the L.P model to determine the quantities of $X$ and $Y$ to be produced,keeping $Z$ in mind, so that the profit earned is maximum.

Formulation of the L.P model:
Let the number of units of products

$X,Y$ and

$Z$ produced be

${x}_{1},{x}_{2},{x}_{z}$ where

${x}_{z}=$ number of units of

$Z$ produced.

$=$ number of units of

$Z$ sold+number of units of

$Z$ destroyed.

$={x}_{3}+{x}_{4}$ (say)
Objective is to maximize the profit.Objective function(profit function) for products

$X$ and

$Y$ is linear because their profits (Rs.

$10$ per unit and Rs.

$20$ per unit) are constants irrespective of the number of units produced.A graph between the total profit and quantity produced will be a straight line.However, a similar graph for product

$Z$ is non-linear since it has slope

$+6$ for first part,while a slope of

$-4$ for the second.However,it is piecewise linear,since it is linear in the regions

$0$ to

$5$ and

$5$ to

$2Y$ .Thus,splitting

${x}_{z}$ in to two parts,viz.the number of units of

$Z$ sold

$\left({x}_{3}\right)$ and number of units of

$Z$ destroyed

$\left({x}_{4}\right)$ makes the objective function for product

$Z$ also linear.
Thus, the objective function is
maximize

$Z=10{x}_{1}+20{x}_{2}+6{x}_{3}-4{x}_{4}.$
Constraints are
on the time available on operation

$I: 3{x}_{1}+4{x}_{2}\le20$
on the time available on operation

$II: 4{x}_{1}+5{x}_{2}\le26$
on the number of units of product

$Z$ sold

$:{x}_{3}\le5$
on the number of units of product

$Z$ produced

$:2Y=Z$
or

$2{x}_{2}={x}_{3}+{x}_{4}$ or

$-2{x}_{2}+{x}_{3}+{x}_{4}=0,$
where

${x}_{1},{x}_{2},{x}_{3},{x}_{4},$ each

$\ge0$

An aeroplane can carry a maximum of 250 passengers, A profit of Rs. 1500 is made on each executive class ticket and a profit of Rs. 900 is made on each economy class ticket. The airline reserves at least 30 seats for executive class. However at least 4 times as many passengers prefer to travel by economy class than by executive class. Formulate LPP in order to maximize the profit For the airline.

Maximum number of passengers

$=250$

Profit on executive class ticket

$=Rs\ 1500$

Profit on Economy class ticket

$=Rs\ 900$

Seats for executive class

$\ge 30$

Numbers of passengers on Executive class

$=x$

Numbers of passengers on economy class

$=y$

$x\ge 30$

$y\ge 4x$

Given

$x+y=250$

$x=250-y$

$\le 250-4x$

$4x\le 250$

$x\le 50$ then

$y\ge 200$

Profit

$=1500 x+900 y$

Maximize

$P$

$P=1500 x+900 y$

$=1500\times 50+900\times 200$

$=75000+180000$

$=255,000$

One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining $15$ camels were seen on the bank of a river. Find the total number of camels.

Let the total number of camels be

$=x$ .

Then number of camels seen in the forest

$=\cfrac { x }{ 4 }$
Number of camels gone mountains

$=2\sqrt {x}$
Number of camels on the river bank

$=15$
Total number of camels

$=\cfrac { x }{ 4 } +2\sqrt { x } +15\quad$

By hypothesis, we have

$\therefore \cfrac { x }{ 4 } +2\sqrt { x } +15=x\quad$

$\Rightarrow 3x-8\sqrt { x } -60=0\quad \quad \Rightarrow 3x-8\sqrt { x } -60=0$

where

$x={y}^{2}$

$\Rightarrow 3{ y }^{ 2 }-8y-60=0\quad \Rightarrow -18y+10y-60=0\Rightarrow 3y(y-6)+10(y-6)=0$

$\Rightarrow (3y+10)(y-6)=0\Rightarrow y=6\quad or\quad y=-\cfrac { 10 }{ 3 }$

Now,

$\quad y=-\cfrac { 10 }{ 3 } \Rightarrow x={ \left( -\cfrac { 10 }{ 3 } \right) }^{ 2 }=\cfrac { 100 }{ 9 }$

But, the number of camels cannot be a fraction

$y=6\Rightarrow x={ 6 }^{ 2 }=36$

Hence, the number of camels

$=36$

A firm produces three products. These products are processed on three different machines. The time required to manufacture one unit of each of the three products and the daily capacity of the three machines are given in the table below:
Machine Time per unit
(in minutes)
of product- $1$ Time per unit
(in minutes)
of product- $2$ Time per unit
(in minutes)
of product- $3$ Machine capacity
(minutes/day)
${M}_{1}$ $2$ $3$ $2$ $440$
${M}_{2}$ $4$ - $3$ $470$
${M}_{2}$ $2$ $5$ - $430$
It is required to determine the daily number of units to be manufactured for each product. The profit per unit for product $1,2$ and $3$ is Rs. $4,$ Rs. $3$ and Rs. $6$ respectively. It is assumed that all the amounts produced are consumed in the market. Formulate the mathematical(L.P) model that will maximize the daily profit.

Formulation of Linear Programming Model:
Step-1:
From the study of the situation find the key decision to be made.In this connection,looking for variables helps considerably.In the given situation key decision is to decide the number of units of products

$1,2$ and

$3$ to be produced daily.
Step-2:
Assume symbols for variable quantities noticed in step-1.Let the number of units of products,

$1,2,$ and

$3$ manufactured daily be

${x}_{1},{x}_{2}$ and

${x}_{3}$ .
Step-3:Express the feasible alternatives mathematically in terms of variables.Feasible alternatives are those which are physically,economically and financially possible.In the given situation feasible alternatives are sets of values of

${x}_{1},{x}_{2}$ and

${x}_{3}$ .
where

${x}_{1},{x}_{2},{x}_{3}\ge0$ .
since negative production has no meaning and is not feasible.
Step-4:Mention the objective quantitatively and express it as a linear function of variables.In the present situation,objective is to maximize and profit.
i.e.,maximize

$Z=4{x}_{1}+3{x}_{2}+6{x}_{3}$
Step-5:Put into words the influencing factors or constraints.These occur generally because of constraints on availability(resources) or requirements(demands).Express these constraints also as linear equations/inequalities in terms of variables.
Here,constraints are on the machine capacities and can be mathematically expressed as

$2{x}_{1}+3{x}_{2}+2{x}_{3}\le440$

$4{x}_{1}+0.{x}_{2}+3{x}_{3}\le470$

$2{x}_{1}+5{x}_{2}+0.{x}_{3}\le430$

$\therefore,$ the complete mathematical(L.P) model for the problem can be written as
Maximize

$Z=4{x}_{1}+3{x}_{2}+6{x}_{3}$
subject to constraints,

$2{x}_{1}+3{x}_{2}+2{x}_{3}\le440$

$4{x}_{1}+3{x}_{3}\le470$

$2{x}_{1}+5{x}_{2}\le430$

A firm produces an alloy having the following specifications:
$\left(i\right)$ specific gravity $\le0.98$
$\left(ii\right)$ chromium $\ge8$ percent
$\left(iii\right)$ melting point $\ge{450}^{0}$ C
Raw materials $A,B$ and $C$ having the properties shown in the table can be used to make the alloy.
Property Property of raw
material A Property of raw
material B Property of raw
material C
Specific gravity $0.92$ $0.97$ $1.04$
Chromium $7$ percent $13$ percent $16$ percent
Melting point ${440}^{0}$ Celsius ${490}^{0}$ Celsius ${480}^{0}$ Celsius
Costs of the various raw materials per ton are:Rs. $90$ for $A$ ,Rs. $280$ for $B$ and Rs. $40$ for $C$ . Formulate the L.P model to find the proportions in which $A,B$ and $C$ be used to obtain an alloy of desired properties while the cost of raw materials is minimum.

Formulation of Linear Programming Model:
Let the proportions of raw materials

$A,B$ and

$C$ to be used for making the alloy be

${x}_{1},{x}_{2}$ and

${x}_{3}$ respectively.
Objective is to minimize the cost
i.e., minimize

$Z=90{x}_{1}+280{x}_{2}+40{x}_{3}$
Constraints are imposed by the specifications required for the alloy.
They are

$0.92{x}_{1}+0.97{x}_{2}+1.04{x}_{3}\le 0.98,$

$7{x}_{1}+13{x}_{2}+16{x}_{3}\ge8,$

$440{x}_{1}+490{x}_{2}+480{x}_{3}\ge 450$ ,
and

${x}_{1}+{x}_{2}+{x}_{3}=1,$
as

${x}_{1},{x}_{2}$ and

${x}_{3}$ are the proportions of materials

$A,B$ and

$C$ in making the alloy.
Also,

${x}_{1},{x}_{2},{x}_{3},$ each

$\ge0$

A company has two grades of inspectors, $I$ and $II$ to undertake quality control inspection.At least $1,500$ pieces must be inspected in an $8-$ hour day.Grade $I$ inspector can check $20$ pieces in an hour with an accuracy of $96$ percent.
Grade $II$ inspector checks $14$ pieces an hour with an accuracy of $92$ percent.
Wages of grade $I$ inspector are Rs. $5$ per hour while those of grade $II$ inspector are Rs. $4$ per hour. Any error made by an inspector costs Rs. $3$ to the company.If there are,in all, $10$ grade $I$ inspectors and $15$ grade $II$ inspectors in the company,find the optimal assignment of inspectors that minimizes the daily inspection cost.

Formulation of L.P model:
Let

${x}_{1}$ and

${x}_{2}$ denote the number of grade

$I$ and grade

$II$ inspectors that may be assigned the job of quality control inspection.
The objective is to minimize the daily cost of inspection.Now the company has to incur two types of costs:wages paid to the inspectors and the cost of their inspection errors.The cost of a grade

$I$ inspector/hour is
Rs.

$\left(5+3\times0.04\times20\right)=$ Rs.

$7.40$
Similarly,cost of a grade

$II$ inspector/hour is
Rs.

$\left(4+3\times0.08\times14\right)=$ Rs.

$7.36.$

$\therefore$ The objective function is
minimize

$Z=8\left(7.40{x}_{1}+7.36{x}_{2}\right)=59.20{x}_{1}+58.88{x}_{2}.$
Constraints are
on the number of grade

$I$ inspectors

$:={x}_{1}\le10,$
on the number of grade

$II$ inspectors

$:={x}_{2}\le15,$
on the number of pieces to be inspected daily

$:20\times8{x}_{1}+14\times8{x}_{2}\ge1,500$
or

$160{x}_{1}+112{x}_{2}\ge1,500,$
where

${x}_{1},{x}_{2}\ge0$

A person wants to decide the constituents of a diet which will fulfill his daily requirements of proteins,fats and carbohydrates at the minimum cost.The choice is to be made from four different types of foods.The yields per unit of these foods are given in table below.
Food Type Proteins
(yield per unit) Fats
(yield per unit) Carbohydrates
(yield per unit) Cost per unit
(in Rupees)
$1$ $3$ $2$ $6$ $45$
$2$ $4$ $2$ $4$ $40$
$3$ $8$ $7$ $7$ $85$
$4$ $6$ $5$ $4$ $65$
Minimum
Requirement $800$ $200$ $700$
Formulate the linear programming model for the problem.

Formulation of L.P. Model
Let

${x}_{1},{x}_{2},{x}_{3}$ and

${x}_{4}$ denote the number of units of type

$1,2,3,$ and

$4$ respectively.
Objective is to minimize the cost i.e.,
Minimize

$Z=$ Rs.

$\left(45{x}_{1}+40{x}_{2}+85{x}_{3}+65{x}_{4}\right)$
Constraints are on the fulfillment of the daily requirements of the various constituents
i.e., for proteins

$3{x}_{1}+4{x}_{2}+8{x}_{3}+6{x}_{4}\ge800,$
for fats

$2{x}_{1}+2{x}_{2}+7{x}_{3}+5{x}_{4}\ge200,$ and
for carbohydrates

$6{x}_{1}+4{x}_{2}+7{x}_{3}+4{x}_{4}\ge700,$
where

${x}_{1},{x}_{2},{x}_{3},{x}_{4}$ each

$\ge0$

Solve the following system of inequalities graphically $2x+y\ge 6, 3x+4y\le 12$

The set of inequalities are given as,

$2x + y \ge 6$

$3x + 4y \le 12$

Simplifying $2x + y \ge 6$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 6$

When $y = 0$ , then, $x = 3$

So, the coordinates are $\left( {0,6} \right)$ and $\left( {3,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \ge 6$

Which cannot be true, so the graph will be shaded away the origin.

Simplifying $3x + 4y \le 12$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 3$

When $y = 0$ , then, $x = 4$

So, the coordinates are $\left( {0,3} \right)$ and $\left( {4,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 12$

Which is true, so the graph will be shaded towards the origin.

1004146_1064086_ans_624ea8f9ec144db4aa394317610a1541.png

Maximize $Z=x+2y$ subject to $2x+y\ge 3,x+2y\ge 6, x,y\ge 0$ .

minimize ,

$Z=x+2y \rightarrow$ objection function

Constraints:-

$2x+y\geq 3$ First draw the graph of

$x+2y\ geq 6$

$\Rightarrow 2x+y=3,x+2y=6$

$x,y \geq 0$

$x=0,y=0$

and shade region according constraints.

Required region = shaded region above BC.

objecting function , z will minimum at the comes i.e

$(03)$ &(6,0)$$

$Z_B=x_B+2y_B=6+2 \times 0=6$ $$

$Z_C=x_c+2y_c=0+2\times 3=6$

$Z=6$

It is clear that the required that the required solution of given L.P.P every point on the line BC and minimum value of

$z=6$

2110476_1040018_ans_34b2ca78cae440658e580178648c8277.png

Draw the line $2y=3x+4$ on the $xy$ plane.

1328612_1054740_ans_3fb240338126477fa87883ddfdec6750.jpg

Solve the following system of inequalities graphically $x+y\ge 4, 2x-y> 12$

The set of inequalities are given as,

$x + y \ge 4$

$2x - y > 12$

Simplifying $x + y \ge 4$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 4$

When $y = 0$ , then, $x = 4$

So, the coordinates are $\left( {0,4} \right)$ and $\left( {4,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \ge 4$

Which cannot be true, so the graph will be shaded away the origin.

Simplifying $2x - y > 12$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = - 12$

When $y = 0$ , then, $x = 6$

So, the coordinates are $\left( {0, - 12} \right)$ and $\left( {6,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 > 12$

Which cannot be true, so the graph will be shaded away the origin.

1004145_1064087_ans_1e79a8af17464d01a3688730945bcb04.png

Solve the following system of inequalities graphically $x+y\le 9, y>x, x\ge 0$

The set of inequalities are given as,

$x + y \le 9$

$y > x$

$x \ge 0$

Simplifying $x + y \le 9$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 9$

When $y = 0$ , then, $x = 9$

So, the coordinates are $\left( {0,9} \right)$ and $\left( {9,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 9$

Which is true, so the graph will be shaded towards the origin.

Now, when $y > x$ , then at $\left( {0,0} \right)$ , $0 > 0$ will give the false statement, so the graph will be away the origin.

1004138_1064094_ans_90abf5fe5aab4571b21c3b1a576d7dd1.png

Solve the following system of inequalities graphically $5x+4y\le 20, x\ge 1, y\ge 2$

The set of inequalities are given as,

$5x + 4y \le 20$

$x \ge 1$

$y \ge 2$

Simplifying $5x + 4y \le 20$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 5$

When $y = 0$ , then, $x = 4$

So, the coordinates are $\left( {0,5} \right)$ and $\left( {4,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 20$

Which is true, so the graph will be shaded towards the origin.

Now, $x = 1$ will give a straight line and since $0 \ge 1$ is not true, so the graph will be shaded away the origin.

In the same way, $y = 2$ will give a straight line and since $0 \ge 2$ is not true, so the graph will be shaded away the origin.

1004135_1064095_ans_4226eb3023a2458f82e134c4c929ef71.png

Solve the following system of inequalities graphically $2x-y>1, x-2y<-1$

The set of inequalities are given as,

$2x - y > 1$

$x - 2y < - 1$

Simplifying $2x - y > 1$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = - 1$

When $y = 0$ , then, $x = 0.5$

So, the coordinates are $\left( {0, - 1} \right)$ and $\left( {0.5,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 > 1$

Which cannot be true, so the graph will be shaded away the origin.

Simplifying $x - 2y < - 1$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 0.5$

When $y = 0$ , then, $x = - 1$

So, the coordinates are $\left( {0,0.5} \right)$ and $\left( { - 1,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 < - 1$

Which cannot be true, so the graph will be shaded away the origin.

1004142_1064089_ans_2c4787e5ee3843d09b168382029afcce.png

Solve the following system of inequalities graphically $2x+y\ge 8, x+2y\ge 10$

The set of inequalities are given as,

$2x + y \ge 8$

$x + 2y \ge 10$

Simplifying $2x + y \ge 8$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 8$

When $y = 0$ , then, $x = 4$

So, the coordinates are $\left( {0,8} \right)$ and $\left( {4,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \ge 8$

Which cannot be true, so the graph will be shaded away the origin.

Simplifying $x + 2y \ge 10$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 5$

When $y = 0$ , then, $x = 10$

So, the coordinates are $\left( {0,5} \right)$ and $\left( {10,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \ge 10$

Which cannot be true, so the graph will be shaded away the origin.

1004139_1064092_ans_e85d68aed0b940889663578e5c316ef1.png

Solve the following system of inequalities graphically $x+y\le 6, x+y\ge 4$

The set of inequalities are given as,

$x + y \le 6$

$x + y \ge 4$

Simplifying $x + y \le 6$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 6$

When $y = 0$ , then, $x = 6$

So, the coordinates are $\left( {0,6} \right)$ and $\left( {6,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 6$

Which is true, so the graph will be shaded towards the origin.

Simplifying $x + y \ge 4$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 4$

When $y = 0$ , then, $x = 4$

So, the coordinates are $\left( {0,4} \right)$ and $\left( {4,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \ge 4$

Which cannot be true, so the graph will be shaded away the origin.

1004140_1064090_ans_1b5f9a4a0130472d808384fa66c97f4a.png

Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, there years from now, I shall be three times as old as you will be. If the present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.

If present age of Aarushi = x years

and present age of Ravish = y years

Their ages seven years ago will be

Aarushi

$= (x - 7)$ years

Ravish

$= (y - 7)$ years

Also their ages three year from now will be

Aarushi

$= (x + 3)$ years

Ravish

$= (y + 3 )$ years

According to question

$y- 7 = 7(x - 7)$ and

$y + 3 = 3 (x + 3)$

$\Rightarrow y - 7 = 7x - 49$

$y + 3 = 3x + 9$

$\Rightarrow 7x - y - 42 = 0$

$3x - y + 6= 0$

For representing graphically task will be

$x$	$7$	$8$	$9$
$y$	$7$	$14$	$21$

$x$	$1$	$2$	$3$
$y$	$9$	$12$	$15$

1523853_1086162_ans_ca4bd8b4bfc440fbb653477709904858.png

Solve the following system of inequalities graphically $3x+4y\le 60, x+3y\le 30, x\ge 0, y\ge 0$

The set of inequalities are given as,

$3x + 4y \le 60$

$x + 3y \le 30$

Simplifying $3x + 4y \le 60$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 15$

When $y = 0$ , then, $x = 20$

So, the coordinates are $\left( {0,15} \right)$ and $\left( {20,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 60$

Which is true, so the graph will be shaded towards the origin.

Simplifying $x + 3y \le 30$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 10$

When $y = 0$ , then, $x = 30$

So, the coordinates are $\left( {0,10} \right)$ and $\left( {30,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 30$

Which is true, so the graph will be shaded towards the origin.

1004132_1064096_ans_b72e12884a2f4a5c963fce403e4baab3.png

If the point $(2K-3,K+2)$ lies on the graph of equation $2x+3y+15=0$ , find the value of k.

Given point

$(2K-3,K+2)$

$2x+3y+15=0$

$2(2K-3)+3(K+2)+15=0$

$4K-6+3K+6+15=0$

$7K+15=0$

$K=-15/7$

Solve the following Linear programming problems graphically :
Maximize $z = 3x + 4y$ subject to the constraints : $x + y \le 4, \, x \ge 0 , \, y \ge 0$

$x+y\leq 4$

x	0	4
y	4	0

Corner points	Value of z=3x+4y
(0,4)	16
(4,0)	9
(0,0)	0

Hence, z is maximum at (0,4) and max value is $16$

See the given figure and write the following:
$(i)$ The coordinates of $B$
$(ii)$ The coordinates of $C$
$(iii)$ The point identified by the coordinates $(-3,-5)$
$(iv)$ The ordinate of the point $B$ .

$(i) B=(-5,2)$

$(ii) C=(5,-5)$

$(iii) (-3,-5)=E$

$(iv)$ Ordinate of

$B=2$

Maximize $Z=-x+2y$ , subject to the constraints: $x\geq 3; x+y\geq 5, x+2 y\geq 6, y\geq 0$ .

1340180_1083064_ans_e2e620aa380d4abcba95bf28398a1bb6.jpg

Solve the following system of inequalities graphically $2x+y\ge 4, x+y\le 3, 2x-3y\le 6$

The set of inequalities are given as,

$2x + y \ge 4$

$x + y \le 3$

$2x - 3y \le 6$

Simplifying $2x + y \ge 4$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 4$

When $y = 0$ , then, $x = 2$

So, the coordinates are $\left( {0,4} \right)$ and $\left( {2,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \ge 4$

Which can never be true, so the graph will be shaded away the origin.

Simplifying $x + y \le 3$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 3$

When $y = 0$ , then, $x = 3$

So, the coordinates are $\left( {0,3} \right)$ and $\left( {3,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 3$

Which is true, so the graph will be shaded towards the origin.

Simplifying $2x - 3y \le 6$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = - 2$

When $y = 0$ , then, $x = 3$

So, the coordinates are $\left( {0, - 2} \right)$ and $\left( {3,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 6$

Which is true, so the graph will be shaded towards the origin.

1004129_1064097_ans_3e8bf57250fe4b028a89dec8c53eaf80.png

Solve the following Linear Programming problems graphically:
Maximize $Z = 3x + 4y$ Subject to the constraints : $x + y \le 4, \, x \ge 0 , \, y \ge 0$
Minimize $Z = -3x + 4 y$ subject to $x + 2y \le 8 , \, 3x + 2y \le 12, \, x \ge 0 , \, y \ge 0$

1.Maximize $Z=3x+4y$

Corner points	Value of $Z=3x+4y$
(0,4)	16
(4,0)	9
(0,0)	0

Hence, Z is maximum at $(0,4)$ and max value is $16$

2.Minimize $Z=-3x+4y$

Corner points	Value of $Z=-3x+4y$
(0,4)	16
(2,3)	6
(4,0)	-12

Hence, Z is minimum at $(4,0)$ and min value is $-12$

$x-y\leq -1,x+y\leq 0,x,y\geq0$

REF.Image.

$x-y\leqslant -1, x+y\leqslant 0, x,y\geqslant 0$

$x-y=1$

x	0	1
y	-1	0

$x+y=0$

x	0	1
y	0	-1

$\therefore$ It is given by

$ar(\triangle AOB)$

$x+y = 0$

$x-y = 1$

$\Rightarrow x = 1+y$

$\Rightarrow 1+y+y = 0$

$y = \frac{-1}{2}$

$\therefore$ Req area =

$ar (\triangle AOB)$

$= \frac{1}{2}\times 1\times \frac{1}{2}$

$\Rightarrow x-1-\frac{-1}{2} = \frac{1}{2}$

$B(\frac{1}{2}, \frac{1}{2})$

$= \frac{1}{4}\, sq. units$

1179393_1201459_ans_841361b781ba49eeaafbd98a98f77cd1.jpg

If a $20$ year old girl drives her car at $25\ km/h$ , she has to spend $Rs\ 4/km$ on petrol. If she drives her car at $40km/h$ , the petrol cost increases to $Rs\ 5\ /km$ . She has $Rs\ 200$ to spend on petrol and wishes to find the maximum distance she can travel within one hour. Express the above problem as Linear Programming Problem. Write any one value reflected in the problem.

Let the distance covered with speed of 25kmph be 'x' km

and the distance covered with speed of 40kmph be 'y' km

Total distance covered be 'z' km

The L.P.P of the above problem is

Maximize

$z=x+y$

Subject to constraints

$4x+5y\le200$

$\dfrac{x}{50}+\dfrac{y}{40}\le1$

$x\ge0,\ y\ge0$

A store sells two types of toys, A and B. The store owner pays Rs. 8 and Rs. 14 for each unit of toy A and toy B respectively. One unit of toy A yields a profit of Rs. 2 while a unit of toy B yields a profit of Rs.The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than Rs. 20,000 in inventory of these toys. How many units of each type of toys should be stocked in order to maximize his monthly total profit ?

Let x be the total number of toys A and y the number of toys B.

x and y cannot be negative. Hence,

$x \geq 0$ and

$y \geq 0$

The store owner estimates that no more than 2000 toys will be sold every month.

$x+y \leq 2000$

One unit of toy A yields a profit of Rs. 2 while a unit of toy B yields a profit of Rs. 3, hence the total profit P is given by

$P=2x+3y$

The store owner pays Rs. 8 and Rs. 14 for each one unit of toy A and B respectively and he does not plan to invest more than Rs. 20,000 in inventory of these toys.

$8x+14y \leq 20000$

The solution set of the system of inequalities given above and the vertices of the region obtained are shown in the figure.

Vertices of the solution set :

A at (0,0)

B at (0,1429)

C at (1333,667)

D at (2000,0)

Calculate the total profit P at each vertex.

P(A) = 0

P(B) = 4287

P(C) = 4667

P(D) = 4000

The maximum profit is at vertex C with x = 1333 and y = 667.

Hence, the store owner has to have 1333 toys of type A and 667 toys of type B in order to maximize his profit.

1305020_1145341_ans_49758eeb744e4999a46f6357d5a83a5c.png

Find the solution of the following LP problem ''Maximize z=x+ y subject $x + y \leq 1,-3x+y\geq 3,x\geq 0,y\geq 0''$ is.............

Consider the problem

$\begin{array}{l} z=x+y \\ x+y\le 1 \\ -3+y\ge 3 \\ x\ge 0,y\ge 0 \end{array}$

And

$\begin{array}{l} x+y=1 \\ \Rightarrow \frac { x }{ 1 } +\frac { y }{ 1 } =1 \end{array}$

points are

$A(1,0)$ ,

$B(0,1)$

And

$\begin{array}{l} -3x+y=3 \\ \Rightarrow \frac { x }{ { -1 } } +\frac { y }{ 3 } =1 \end{array}$

points are

$C(-1,0)$ ,

$D(0,3)$

Corner points are

$(0,0),(0,1),(1,0)$

$z$ at

$(0,0)=0$

$z$ at

$(1,0)=1$

$z$ at

$(0,1)=1$

So, Two solutions of the given

$L.L.P$ exist

$x=1,y=0$ Or

$x=0,y=1$

1195516_1156703_ans_2c6edd7d0031411d87c447fc49d4c314.PNG

A firm is engaged in producing two models ${X_{1.}}{X_2}$ performing only three operations assembling, pianting and testing.The relevant data are as follows.
Total number of hours available each week for assemblling $600$ hours, painting $100$ hours and testing $30$ hours.The firm wishes to determine its weekly products mix so as to maximize revenue formulate I.P.P model for maximize the revenue

Let the no. of models of types

$X_1,X_2$ be x and y respectively.

According to the table,

Total no. of hours of required for assembling

$=1\times x+1.5\times y=x+1.5y$

Total no. of hours of required for painting

$=0.2\times x+0.2\times y=0.2(x+y)$

Total no. of hours of required for testing

$=0\times x+0.1\times y=0.1y$

According to question,

Maximum no. of hours available for assembling

$=600$ hrs

$\therefore x+1.5y\leq600\Rightarrow 2x+3y\leq1200....(1)$

Maximum no. of hours available for painting

$=100$ hrs

$\therefore 0.2(x+y)\leq100\Rightarrow x+y\leq500....(2)$

Maximum no. of hours available for testing

$=30$ hrs

$\therefore 0.1y\leq30\Rightarrow y\leq300....(3)$

The objective function is the revenue

$Z=50x+80y$ which we have to maximize against the constraints (1),(2) and (3)

Using the graphical method of LPP we first need to plot the equation of constraints and find the feasible region.

According to the graph and inequations of the constraints, the feasible region is the area formed by the line x+y=500

So, the corner points are

$(0,0),(0,250),(250,0)$

Value of Z at (0,0):

$Z=50\times0+80\times0=0$

Value of Z at (0,250):

$Z=50\times0+80\times250=20000$

Value of Z at (250,0):

$Z=50\times250+80\times0=12500$

So Z is maximum at

$x=0$ and

$y=250$

1087094_1199695_ans_b431eb245f9e4ad884a2b1607c8e9a93.png

There are $50$ apples trees in an orchard. Each tree produces $800$ apples. For each additional tree planeted in the orchord. The output additional tree drops by $10$ apples tree drops by $10$ apples. Number of trees that should to the existing orchord for maximising the output of the trees is $3k$ , then $k=$ ?

The output for x no. of bees is than

$=\dfrac{800-10k}{50+k}$

$0=300k-10k^2+40000$

$300-20n=0$

Hence,

$n=15$

Hence,

$15$ trees should be added

$3k=15$

$k=5$ .

1190646_1155542_ans_061fc7b376d74231a7f4d4f42d18ccb0.jpg

A producer has $20$ and $10$ units of labour and capital respectively which he can use to produce two kinds of goods $X$ and $Y$ . To produce one unit of $X,2$ units of capital and $1$ unit of labour is required. To produce one unit of $Y,3$ units of labour and $1$ unit of capital is required. Find $X$ and $Y$ ?

Let capital be

$'a'$ ans Labour be

$='b'$

Total capital present

$=10\ a$

Total labour present

$=20\ b$

One unit of

$X$ requires

$=(2a+b)$

One unit of

$Y$ requires

$=(a+3b)$

no. of

$X$ and

$Y$ he can product at atoms

$pX+qY=Total\ capital + Total\ labour$

$pX+qY=10\ a+20\ b$

$\Rightarrow \ p(2a+b)+q(a+3b)=10\ a+20\ b$

$\Rightarrow \ (2p+q)a+(p+3q)6=10\ a+20\ b$

On comparing :

$2p+q=10---(1)$

$p+3q=20\Rightarrow 2p+6q=40----(2)$

$(2)-(1)$

$5q=30$

$\Rightarrow \ q=6 ; 2p+6=10$

$\Rightarrow \ p=2$

$\therefore \$ The products on produced

$2$ units of

$X$ and

$6$ units of

$Y$ at a given time.

Maximize $z=3x+5y$ , Subject to $x+4y \le 24$ , $3x+y \le 21,x+y \le 9,x \ge 0$ . Also find maximum value of $z$ .

Shaded portion OABCD is the feasible region,

Where

$O(0,0),\ A(7,0),\ D(0,6)$

For B:

$3x+y=21....(i)$

$x+y=9......(ii)$

Subtract (ii) from (i) we get

$3x+y-x-y=21-9$

$2x=12$

$x=6$

$y=9-6=3$

$\therefore B(6,3)$

For C:

$x+y=9....(iii)$

$x+4y=24......(iv)$

Subtract (ii) from (i) we get

$x+y-x-4y=9-24$

$-3y=-15$

$y=5$

$x=9-5=4$

$\therefore C(4,5)$

$Z=3x+5y$

$Z\ at\ O(0,0)=3(0)+5(0)=0$

$Z\ at\ A(7,0)=3(7)+5(0)=21$

$Z\ at\ B(6,3)=3(6)+5(3)=33$

$Z\ at\ C(4,5)=3(4)+5(5)=37$

$Z\ at\ D(0,6)=3(0)+5(6)=30$

Thus, Z is maximized at C(4,5) and its maximum value is 37.

1036683_1141551_ans_450c0511dac148d883b356bbe8b3573a.png

A small firm makes necklaces and bracelets. The combined number of necklaces and bracelets that it can handle per day is at most $24$ . A bracelet takes one hour to make and necklace takes $30$ minutes. The maximum number of hours available per day is $16$ . If the profit on a bracelet is Rs. $300$ /- and on a necklace is Rs. $100$ /-, How many necklaces and bracelets must be made per day to get the maximum profit? Make it as an LLP and solve graphically.

Suppose a number of necklaces manufacture be

$x$ and the number of bracelets be

$y$ .

Since total number of items area at most

$24$

$x+y\le 24.....(1)$

Bracelets takes

$1$ hour to manufacture and necklaces take half an hour to manufacture.

$x$ items takes

$x$ hours to manufacture and if items take

$y/2$ hour to manufacture and minimum time available is

$16$ hours

$\therefore$

$x/2+y\le 16.....(2)$

The profit one necklace is Rs.

$100$ and the profit one bracelet is Rs.

$300$

Let the profit be

$t$ . Now we wish to maximise the profits so

$max$

$z=100x+300y....(3)$

So,

$x+y\le 24$

$\cfrac{x}{2}+y\le 16$

max

$z=100x+300y$ is required LLP

Maximize z = 2x + 4y subject to the constraints

x ≥ 0, y ≥ 0, x + y ≥ 1, 3x + 2y <= 6

$x\le0$

When

$x$ is greater than zero it mean tangent of that mean the feasible region is on or to the right of the line where equation is zero

$(x=0)$ which is the line which is

$y$ axis

$y\le0$

Where y is greater that it means above that mean feasible region is on or above the line where equation is

$y=0$ which is the line which is x-axis

Those two mean that the feasible region is in the upper right hand part of the xy co ordinate system.

See image 1

When x is greater that it mean right of

$y\le 1-x$

That mean feasible region os on or of above the line where equation is

$x+y=1$

See image 2

$3x+2y\le6$

If we solve

$x\le\cfrac{6-2y}{3}\\ y\le\cfrac{6-3x}{2}$

$x \quad intercept (2,0)\quad\quad y\quad intercept (0,3)\\ Corner \quad point\quad\quad\quad value \quad of\quad 2=2x+4y$

$(1,0)\quad \quad \quad \quad \quad \quad \quad 2=2(1)+4(10)=2+0=2\\ (2,0)\quad \quad \quad \quad \quad \quad \quad 2=2(2)+4(0)=4+0=4\\ (0,3)\quad \quad \quad \quad \quad \quad \quad 2=2(0)+4(3)=0+12=12\\ (0,1)\quad \quad \quad \quad \quad \quad \quad 2=2(0)+4(1)=0+4=4$

Maximum value when

$x=0\quad y=3$

1211326_1204670_ans_9e7dc2f652484ceea408f9f97eea65e0.png

Maximise ,Z=x+y, subject to $x-y\leq -1$ , $-x+y\leq 0$ $x,y\leq 0$

$\begin{array}{l} Maximize \\ z=\left( { x+y } \right) \\ \Rightarrow x-y\le -1,y-x\le 0,x,y\le 0 \\ \Rightarrow put\, \, x=0\, \, in\, \, x-y=-1\to \left( i \right) \\ y=1 \\ \Rightarrow put\, \, x=1\, \, in\, \, equation\left( i \right) \\ y=2 \\ x=-1 \\ -1-y=-1 \\ y=0 \\ and\, \, in\, \, y-x=0 \\ \Rightarrow put\, \, x=0 \\ y=0 \\ \Rightarrow put\, x=1 \\ y=1 \\ \Rightarrow put\, x=2 \\ y=2 \\ Put\, \, x=-1 \\ y=-1 \end{array}$

No feasible region, hence, no maximum value

1221050_1211771_ans_056692b4e73a4cbb8aa072dc3712f325.JPG

A library has to accommodate two different types of books on a shelf. The books are 6cm and 4cm thick and weigh 1kg and 1.5kg respectively. The shelf is 96cm long and at most can support a weight of 21kg. How should the shelf be filled with the books of two types in order to include the greatest number of books? Make it as an LLP and solve it graphically.

Consider the problem

Let two type of books be

$x$ and

$y$ ,

The required LLP is maximize

$Z=x+y$ Subject to constraints

$\begin{array}{l} 6x+4y\le 96\, Or\, 3x+2y\le 48 \\ x+\frac { 3 }{ 2 } y\le 21\, Or\, 2x+3y\le 42 \end{array}$

and

$x,y \ge 0$

On considering the inequalities as equations, We get

$\begin{array}{l} 3x+2y=48\, \, ...\left( i \right) \\ 2x+3y=42\, ....\left( { ii } \right) \end{array}$

Now tablefor line

$3x + 2y = 48$ is

x	0	16
y	24	0

So, it passes through

$(0,0)$ and

$(16,0)$

On putting

$(0,0)$ in

$3x + 2y \le 48$ we get

$\begin{array}{l} 0+0\le 48 \\ Or\, 0\le 48\, \, \, \left[ { which\, is\, true } \right] \end{array}$

so, the half plane is towards the origin.

And Table for

$2x + 3y = 42$ is

x	0	21
y	14	0

So, it passes through

$(0,14)$ and

$(21,0)$ .

On putting

$(0,0)$ in

$2x + 3y \le 42$ We get

$\begin{array}{l} 0+0\le 42 \\ Or\, 0\le 42\, \, \left[ { which\, is\, true } \right] \end{array}$

On solving equation

$(i)$ and

$(ii)$ we get

$x=12$ and

$y=6$

Thus, the point of intersection is

$B(12,6)$

And from the graph

$OABCD$ is the feasible region which is bounded. The corner points are

$O(0,0),A(0,14),B(12,16),C(16,0)$ .

And the value of

$Z$ at corner points are

Corner points	Value of $Z=x+y$
$O(0,0)$	$Z=0+0=0$
$A(0,14)$	$Z=0+14=14$
$B(12,16)$	$Z=12+6=18(maximum)$
$C(16,0)$	$Z=16+0=16$

From the table the maximum value of

$Z$ is

$18$ at

$B(12,6)$ .

Hence, The maximum number of books of

$I$ type is

$12$ and books of

$II$ type is

$6$ .

1193788_1212047_ans_fae5aa9bdb5e4006b226e3584830e4d5.PNG

A small firm manufactures gold rings and chains.The total number of rings and chains manufactured per day is almost $24$ . It takes $1\ h$ to make a ring and $30$ min to make a chain. The maximum number of hours available per day is $16$ . If the profit on a ring is $Rs. 300$ and that on a chain is $Rs. 190$ , then find the number of rings and chains that then find the number of rings and chains that should be manufactured per day so as to earn the maximum profit. Make it as an LPP and solve it graphically.

intersection point of x+y=24 and 2x+y=32

$2x+y=32$ ......................(i)

$x+y=24$ ....................(ii)

subtract (ii) from (i)

$x=8$

Now put x in(ii)

$y=16$

1672256_1473055_ans_c756c5955ae341ed9cf7d255a0609379.jpeg

The standard weight of a special purpose brick is $5\mathrm { kg }$ and It must contain two basic ingredients $B _ { 1 } \text { and } B _ { 2 } , B _ { 1 }$ costs $Rs. 5$ per kg and $B _ { 2 }$ costs $Rs.8$ per kg, Strength considerations dictate that the brick should contain not more than $4\mathrm { kg }$ of $B _ { 1 }$ and minimum $2 \mathrm { kg }$ of $\mathrm { B } _ { 2 } .$ Since the demand for the product is likely to be related to the price of the brick. find the minimum cost of brick satisfying the above conditions, Formulate this situation as an $LPP$ and solve it graphically.

1961047_1481940_ans_427fbfdf18a84aae85889c0381878940.jpg

A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires $3g$ of silver and $1$ g of gold while that of B requires $1$ g of silver and $2$ g of gold. The company can use atmost $9$ g of silver and $8$ g of gold. If each unit of type A brings a profit of $40$ and that of type B $50$ , find the number of units of each type that the company should produce to maximize the profit. Formulate and solve graphically the LPP and find the maximum profit.

1693223_1517190_ans_1c5c099e979d4aa184d6b8b20b9f017b.jpg

A man has $Rs. 1500$ for purchasing wheat and rice. A bag of rice and a bag of wheat cost $Rs. 180$ and $Rs. 120$ respectively. He has a storage capacity of only $10$ bags. He earns a profit of $Rs. 11$ and $Rs. 9$ per bag of rice and wheat, respectively. Formulate the problem as an $LPP$ to find the number of bags of each type he should buy for getting maximum profit and solve it graphically.

1672255_1447099_ans_21ed2344eb874e9ca23dcf784a23ef1f.jpeg

Maximize $Z=12x+24y$ subject to the constraints $x+y\ge 5, 5x+7y\le 35, x-y\ge0,x,y\ge0$ graphically.

$Z = 12x + 24y \quad \left( \text{Given} \right)$

Subject to the constraints:

$x + y \ge 5$

$5x + 7y \le 35$

$x - y \ge 0$

$x, y \ge 0$

Case I:-

$x + y \ge 5$

$x$	$0$	$5$
$y$	$5$	$0$

Case II:-

$5x + 7y \le 35$

$x$	$0$	$7$
$y$	$5$	$0$

Case II:-

$x - y \ge 0$

$x$	$0$	$1$
$y$	$0$	$1$

$Max \; Z = 12x + 24y$

Corner points	Value of $Z$
$\left( \cfrac{5}{2}, \cfrac{5}{2} \right)$	$90$
$\left( 2.9, 2.9 \right)$	$104.4$
$\left( 5, 0 \right)$	$60$
$\left( 7, 0 \right)$	$84$

$Max \; Z = 104.4$ when

$x = y = 2.9$ .

Hence the maximum value of

$Z$ is

$104.4$ .

1211698_1512665_ans_36afff103a1e4419b38651a4232033d6.jpeg

A firm has to transport at least 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost of engaging each large van is Rs. 400 and each small van is Rs.Not more than Rs. 3000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost.

1692270_1367691_ans_f81c88db5876435e8e869a21f737a6bd.jpg

One kind of cake requires $300gm$ of flour and $15gm$ of fat and another kind of cake requires $150gm$ of flour and $30gm$ of fat. Find the maximum number of cake that can be made from $7.5kg$ of flour and $600gm$ of fat . Form a linear programming problem and solve it graphically.

Let

$x$ and

$y$ be the number of cake of first and second kind respectively.

Case I:- Given that first kind of cake requires

$300 \; g$ of flour while the second kind of cake required

$150 \; g$ of flour.

Maximum quantity available

$= 7.5 \; kg = 7500 \; g$

Therefore,

$300 x + 150 y \le 7500$

$2x + y \le 50; \quad x \ge 0, \; y \ge 0$

$x$	$0$	$25$
$y$	$50$	$0$

Case II:- Given that first kind of cake requires

$15 \; g$ of fat while second kind of cake required

$30 \; g$ of fat.

Maximum quantity available

$= 600 \; g$

Therefore,

$15 x + 30 y \le 600$

$x + 2y \le 40; \quad x \ge 0, \; y \ge 0$

$x$	$0$	$40$
$y$	$20$	$0$

Total no. of cakes

$= x + y$

$Max \; Z = x + y$

Corner points	Value of $Z$
$\left( 0, 20 \right)$	$20$
$\left( 20, 10 \right)$	$30$
$\left( 25, 0 \right)$	$25$

Thus value of

$Z$ is maximum when

$x = 20$ and

$y = 30$ , i.e.,

No. of cakes will be maximum when the no. of cakes of first and second kind are

$20$ and

$10$ respectively.

Hence the maximum no. of cakes that can be made are

$30$ .

1211668_1512681_ans_efe1d8785b034df18cd634ba8fa22bf6.jpeg

A manufacturer of patent medicines is preparing a production plan on medicines $A$ and $B$ . There are sufficient row material available to make $20000$ bottles of $A$ and $40000$ bottles of $B$ , but there are $45000$ bottles into which either of the medicines can be put. Further, it takes $3$ hours to prepare enough material to fill $1000$ bottles of $A$ , it takes $1$ hours to prepare enough material to fill $1000$ bottles of $B$ and there are $66$ hours available for this operation. The profit is Rs. $8$ per bottle for $A$ and Rs. $7$ per bottle for $B$ . Construct the maximization problem.

Let production of each of

$A$ and

$B$ are

$x$ and

$y$ respectively.

Since profits on each bottle of

$A$ and

$B$ are Rs.

$7$ per bottle respectively. So profit on

$x$ bottles of

$A$ and

$y$ bottles of

$B$ are

$8x$ and

$7y$ respectively.

Let

$Z$ be total profit on bottles so,

$Z=8x+7y$ .

Since, it takes

$3$ hours and

$1$ hour to prepare enough material to fill

$1000$ bottles of Type

$A$ and Type

$B$ respectively. so

$x$ bottles of

$A$ and

$y$ bottles of

$B$ are preparing is

$\dfrac{3x}{1000}$ hours and

$\dfrac{y}{1000}$ hours respectively, but only

$66$ hours are available, so

$\dfrac{3x}{1000}+\dfrac{y}{1000}\le 66$

or,

$3x+y\le 66000$ .

Since raw material available to make

$2000$ bottles of

$A$ and

$4000$ bottles of

$B$ but there are

$45000$ bottles in which either of these medicines can be put so,

$x\le 2000$ ,

$y\le 40000$ ,

$x+y\le 45000$ with

$x,y\ge 0$ .

So mathematical formulation of the given LPP is

Max

$Z=8x+7y$

Subject to

$3x+y\le 66000$ ,

$x\le 2000$ ,

$y\le 40000$ ,

$x+y\le 45000$ with

$x,y\ge 0$ .

Maximize $z=6x+4y$ , Subject to $x\leq 2, x+y\leq 3$ , $-2x+y\leq 1$ , $x\geq 0$ , $y\geq 0$ . Also find maximum value of z.

1310642_1360438_ans_e499a0a5eb1d4d96b51aad9444b9790b.jpg

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require $5$ minutes each for cutting and $10$ minutes each for assembling. Souvenirs of type $B$ require $8$ minutes each for cutting and $8$ minutes each for assembling. There are $3$ hours and $20$ minutes available for cutting and $4$ hours available for assembling. The profit is $Rs. 5$ each for type $A$ and $Rs. 6$ each for type $B$ souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit ? Formulate the above $LPP$ and solve it graphically and also find the maximum profit.

Let the company manufacture

$x$ souverins of type

$A$ and

$y$ souvenirs of type

$B$ .
Therefore,

$x\ge 0$ and

$y\ge 0$
The given information can be compiled in a table as follows.

	Type A	Type B	Availability
Cutting (min)	$5$	$8$	$3\times 60+20=200$
Assembling (min)	$10$	$8$	$4\times 60=240$

The profit on type

$A$ souvenirs is Rs.

$5$ and on type

$B$ souvenirs is Rs.

$6$ .Therefore, the constraints are

$5x+8y\le 200$

$10x+8y\le 240$ i.e.,

$5x+4y\le 120$
Total profit,

$Z=5x+6y$
The mathematical formulation of the given problem is
Maximise

$Z=5x+6y.........(1)$
subject to the constraints

$5x+8y\le 200........(2)$

$5x+4y\le 120........(3)$

$x,y\ge 0........(4)$
The feasible region determined by the system of constraints is as shown.
The corner points are

$A(24,0), B(8,20)$ and

$C(0,25)$
The values of

$Z$ at these corner points are as follows.

Corner point	$Z=5x+6y$
$A(24,0)$	$120$
$B(8,20)$	$160$	$\rightarrow$ Maximum
$C(0,25)$	$150$

The maximum value of

$Z$ is

$200$ at

$(8,20)$ .
Thus,

$8$ souvenirs of type

$A$ and

$20$ souvenirs of type

$B$ should be produced each day to get the maximum profit of Rs.

$160$

1951978_1595593_ans_5420d8bb281c49ef82e06194dca1b757.png

If the replacement set is the set of all non-negative integers, find the solution set of the following
A) $x + 1 < 9$
B) $3x - 4 \leqslant 20$
C) $5 + 2x > 10$
D) $10 < x + 5 < 21$

$x+1<9$

$x<9-1\Rightarrow x<8$

$x\in \left[ 0,7 \right]$

$3x-4\le 20$

$3x\le 24\Rightarrow x\le 8$

$x\in \left[ 0,8 \right]$

$5+2x>10$

$2x>5$

$x>\cfrac { 5 }{ 2 } ,x\in \left[ 3,\infty \right]$

$10<x+5<21$

$5<x<16$

$x\in \left[ 6,15 \right]$

An automobile manufacturer makes automobiles and trucks in a factory that is divided into two shops. Shop $A$ , which performs the basic assembly operation, must work $5$ man-days on each truck but only $2$ man-days on each automobile. Shop $B$ , which performs finishing operations, must work $3$ man-days for each automobile or truck that it produces. Because of men and machine limitation, shop $A$ has $180$ man-days per week available whole shop $B$ has $135$ man-days per week. If the manufacturer makes a profit of Rs. $30000$ on each truck and Rs. $2000$ on each automobile, how many of each should he produce to maximum his profit? Formulate this as a $LPP$ .

Let

$x$ number of trucks and

$y$ number of automobiles were produced to maximize the profit.

Since, the manufacturer makes profit of

$Rs.30000$ on each truck and

$Rs.2000$ on each automobile.

$\therefore$ On

$x$ number of trucks and

$y$ number of automobiles profit would be

$Rs.30000x$ and

$Rs.2000y$ respectively.

Total profit

$=Rs.(30000x+2000y)$

Let

$z$ denotes the total profit. Then,

$\Rightarrow$

$z=30000x+2000y$

Since,

$5$ man -days and

$2$ man-days were required to produce each truck and automobile at shop

$A.$

$\therefore$

$5x$ man-days and

$2y$ man-days are required to produce each truck and

$y$ automobiles at shop

$A.$

Also,

Since

$3$ man-days were required to produce each truck and automobile at shop

$B.$

$\therefore$

$3x$ man-days and

$3y$ man-days are required to produce

$x$ trucks and

$y$ automobiles.

A, shop

$A$ has

$180$ man-days per week available while shop

$B$ has

$135$ man-days per week.

$\therefore$

$5x+2y\le 180,$

$3x+3y\le 135$

Number of trucks an automobiles cannot be negative.

$\therefore$

$x,y\ge 0$

Hence, the required LPP is as follows

$:$

Maximize

$z=30000x+2000y$

Subject to

$5x+2y\le 180,$

$3x+3y\le 135,$

$x\ge 0,\,y\ge 0$

A small manufacturing firm produces two types of gadgets $A$ and $B$ , which are first processed in the foundry then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of $A$ and $B$ , and the number of man-hours the firm has available per week are as follows:
Gadget Foundry Machine-shop
$A$ $10$ $5$
$B$ $6$ $4$
Firm's capacity per week $1000$ $600$
The profit of the sale of $A$ is Rs. $30$ per unit as compared with Rs. $.20$ per unit of $B$ . The problem is to determine the weekly production of gadgets $A$ and $B$ , so that the total profit is maximized. Formulate this problem as a LPP.

The given data can be shown in a table as follows :

Gadget	Foundry	Machine shop	Profit
$A$	$10$	$5$	$Rs.30$
$B$	$6$	$4$	$Rs.20$
Fir'ms capacity per week	$1000$	$600$

Now, let the required weekly production of gadgets

$A$ and

$B$ be

$x$ and

$y$ respectively.

As it is given that profit weekly production of gadgets

$A$ is

$Rs.30$ and that on

$B$ is

$Rs.20,$ so profit on

$x$ and

$y$ number of gadgets

$A$ and

$B$ are

$30x$ and

$20y.$

$z=$ Total profit then, we have,

$\Rightarrow$

$z=30x+20y$

It is also given that the production of

$A$ and

$B$ requires

$10$ hours per week and

$6$ hours per week in the foundry. Also, the maximum capacity of the foundry is given as

$1000$ hours.

Now,

$x$ units of

$A$ and

$y$ units of

$B$ will require

$10x+6y$ hours.

So, we have,

$\Rightarrow$

$10x+6y\le 1000$

This is our first constraint.

Given, production of one unit gadget

$A$ requires

$5x$ hours per week and

$y$ units of gadget

$B$ requires

$4y$ hours per week, but the maximum capacity of the machine shop is

$600$ hours per week. So,

$\Rightarrow$

$5x+4y\le 600$

This is our second constraint.

Hence, the mathematical formulation of

$LPP$ is :

Find

$x$ and

$y$ which will maximize

$z=30x+20y$

Subject to constraints,

$\Rightarrow$

$10x+6y\le 1000$

$\Rightarrow$

$5x+4y\le 600$

And also, as production cannot be less than zero, so,

$x,y\ge 0$

Two tailors $A$ and $B$ earn Rs. $300$ and Rs. $400$ per day respectively. A can stitch $6$ shirts and $4$ pants per day while $B$ can stitch $10$ shirts and $4$ pants per day. Form a linear programming problem to minimize the labour cost to produce at least $60$ shirts and $32$ pants.

$Tailors/Product$	$A(x)$	$B(y)$	$Avl$
$Shirts$	$6$	$10$	$60$
$Pants$	$4$	$4$	$32$

Let

$A$ work

$x$ days and

$B$ work$ y$$ days.

$\therefore$

$x\ge 0,$

$y\ge 0$

So, L.P.P., objective function Minimize

$z=300x+400y$

Subject to,

$6x+10y\ge 60$

$4x+4y\ge 32$

Therefore LPP,

Minimize,

$z=300x+400y$

Subject to,

$3x+5y\ge 30$

$x+y\ge 8$

$x\ge 0,$

$y\ge 0$

Vitamin $A$ and $B$ are found in two different foods $F_{1}$ and $F_{2}$ . One unit of food $F_{1}$ contains $2$ units of vitamin $A$ and $3$ units of vitamin $B$ . One unit of food $F_{2}$ contains $4$ units of vitamin $A$ and $2$ units of vitamin $B$ . One unit of food $F_{1}$ and $F_{2}$ cost Rs. $50$ and $25$ respectively. The minimum daily requirement for a person of vitamin $A$ and $B$ is $40$ and $50$ units respectively. Assuming that any things in excess of daily minimum requirement of vitamin $A$ and $B$ is not harmful, find out the optimum mixture of food $F_{1}$ and $F_{2}$ at the minimum cost which meets the daily minimum requirement of vitamin $A$ and $B$ . Formulate this as a $LPP$ .

Let

$x$ and

$y$ units of food

$F_1$ and food

$F_2$ were mixed.

Clearly,

$x\ge 0$ and

$y\ge 0$

One unit of food

$F_1$ contains

$2$ units of vitamin

$A$ and one unit of food

$F_2$ contains

$4$ units of vitamin

$A.$

$\therefore$

$x$ and

$y$ units of food

$F_1$ and food

$F_2$ respectively contains

$2x$ and

$4y$ units of vitamin

$A.$

It is given that the minimum daily requirements for a person of vitamin

$A$ is

$40$ units.

Hence,

$2x+4y\ge 40$

One unit of food

$F_1$ contains

$3$ units of vitamin

$B$ and one unit of food

$F_2$ contains

$2$ units of vitamin

$B.$

$\therefore$

$x$ and

$y$ units of

$F_1$ and

$F_2$ contains

$3x$ and

$2y$ units of vitamin

$B.$

It is given that the minimum daily requirements for a person of vitamin

$B$ is

$50$ units.

Hence,

$3x+2y\ge 50$

One unit of food

$F_1$ and food

$F_2$ cost

$Rs.50$ and

$25$ respectively.

$\therefore$

$x$ and

$y$ units of food

$F_1$ and food

$F_2$ costs

$Rs.50x$ and

$Rs.25y$ respectively.

Let

$z$ denote the total cost

Then,

$z=Rs(50x+25y)$

Hence, the required LPP is

Minimize

$Z=50x+25y$

Subject to

$2x+4y\ge 40$

$3x+2y\ge 50$

$x\ge0,$

$y\ge 0$

Solve each of the following linear programming problem by graphical method.
Minimize $Z=5x+3y$
Subject to $2x+y\ge 10$
$x+3y\ge 15$
$x\le 10$
$y\le 8$
$x, y\ge 0$

1622191_1664208_ans_419e7aa34f3a49668ba366bc828220be.jpg

Solve each of the following linear programming problem by graphical method.
Minimize $Z=18x+10y$
Subject to $4x+y\le 20$
$2x+3y\le 30$
$x,y\ge 0$

1619207_1664156_ans_3e24b3543f1c4de9993e2b1553d0accd.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=7x+10y$
Subject to $x+y\le 30000$
$y\le 12000$
$x\ge 6000$
$x\ge y$
$x, y\ge 0$

1621995_1664199_ans_3e8ea6ae35774f07ad2badf109228878.jpg

Solve each of the following linear programming problem by graphical method.
Minimize $Z=2x+4y$
Subject to $x+y\ge 8$
$x+4y\ge 12$
$x\ge 3, y\ge 2$

1622117_1664203_ans_b1648d356c0d4e0d96c345a23325a48b.jpg

A firm has to transport at least $1200$ packages daily using large vans which carry $200$ packages each and small vans which can take $80$ packages each. The cost of engaging each large and the number of large vans is Rs. $400$ and each small can is Rs. $200$ . Not more than Rs. $3000$ is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost.

1637319_1664100_ans_5241659eaac044b0a2933cc14e572c28.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=15x+10y$
Subject to $3x+2y\le 80$
$2x+3y\le 70$
$x,y\ge 0$

1620512_1664184_ans_6e4fdf1b2d7b4b2aafaeacaace90999a.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=10x+6y$
Subject to $3x+y\le 12$
$2x+5y\le 34$
$x,y\ge 0$

1620663_1664191_ans_e04b3b47fdfb429093e67ee3ecf62ac5.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=9x+3y$
Subject to $2x+3y\le 13$
$3x+y\le 5$
$x,y\ge 10$

1619117_1664155_ans_2384a8544e00409aba056f762026e7ec.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=3x+4y$
Subject to $2x+2y\le 80$
$2x+4y\le 120$

1620660_1664195_ans_86ddd0239f0740ba944d9e34a2d8b549.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=4x+3y$
Subject to $3x+4y\le 24$
$8x+6y\le 48$
$x\le 5$
$y\le 6$
$x, y\ge 0$

1620505_1664179_ans_9ead8e6cdd534b8b94727ed02e79b53b.jpg

A diet for a sick person must contain at least $4000$ units of vitamins, $50$ units of minerals and $1400$ of calories. Two foods $A$ and $B$ , are available at a cost of $Rs. 5$ and $Rs. 4$ per unit respectively. If one unit of $A$ contains $200$ units of vitamins, 1 unit of minerals and 40 units of calories, while one unit of the food B contains 100 unit of vitamins, 2 unit of minerals and 40 units of calories. Find what combination of foods should be used to have the least cost, but it must satisfy the requirements of the sick person. Form the question as LPP and solve it graphically.

1572454_1664255_ans_3bfed26f20004c399d03503eb7bb3a93.jpg

Kellogg is a new cereal formed of a mixture of a mixture of bran and rice that contains at least $88$ grams of proteins and at least $36$ milligrams of iron. knowing that bran contains $80$ grams of protein and $40$ milligrams of iron per kilogram, and that rice constant $100$ grams of protein and $30$ milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs $Rs. 5$ per kg and rice costs $Rs. 4$ per kg.

1573869_1664331_ans_637ae5add9dd49ae972361a5aeb3261d.png

A publisher sells a hard cover edition of a text book for $Rs. 72$ and a paperback edition of the same ext for $Rs. 40$ . Costs to the publisher are $Rs. 56$ and $Rs. 28$ per book respectively in addition to weekly costs of $Rs. 9600$ . Both types requires $5$ minutes of printing time, although hardcover requires $10$ minutes binding time and the paperback requires only $2$ minutes. Both the printing the binding operation have $4,800$ minutes available each week. How many of each type of book should be produced in order to maximize profit?

1573871_1664846_ans_f3d8c8280c1646a795c105df5a3fb21e.png

A gardener has supply to fertilizer of type I which consists of $10\%$ nitrogen and $6\%$ phosphoric acid and type II fertilizer which consists of $5\%$ nitrogen and $10\%$ phosphoric acid. After testing the soil conditions, he finds that he needs at least $14$ kg of nitrogen and $14$ kg of phosporic acid for his crop. If the type I fertilizer costs $60$ paise per kg and type II fertilizer costs $40$ paise per kg, determine how many kilogram of each fertilizer should be used so that nutrient requirement are met at a minimum cost. What is the minimum cost?

1573757_1665198_ans_03f355dc81b84f65803b4d7c0103a067.png

Anil wants to invest at most $Rs. 12000$ in saving certificates and national saving bonds. According to rule, he has to invest at least $Rs. 2000$ in saving certificates and at least $Rs. 4000$ in National Saving Bonds. If the rate of interest on saving certification is $8\%$ per annum and the rate of interest on National Saving Bond is $10\%$ per annum, how much money should he invest to earn maximum yearly income? Also find his maximum yearly income.

1573759_1665202_ans_fccd413359d74fd9a7893a9d83b7c78a.png

Solve each of the following linear programming problem by graphical method.
Maximize $Z=3x+5y$
Subject to $x+2y\le 20$
$x+y\le 15$
$y\le 5$
$x,y\ge 0$

1622273_1664238_ans_68c40481467f4f41ad25b1d9119fda4b.jpg

A firm manufactures two types of products $A$ and $B$ and sells them at a profit of $Rs. 5$ per unit of type $A$ and $Rs. 3\ per$ unit of type $B$ . Each product is processed on two machines $M_{1}$ and $M_{2}$ . One unit of type $A$ requires one minutes of processing time on $M_{1}$ and two minutes of processing time on $M_{2}$ , whereas one unit of type $B$ requires one minutes of processing time on $M_{1}$ and one minutes on $M_{2}$ . Machines $M_{1}$ and $M_{2}$ are respectively available for at most $5$ hours and $6$ hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.

1573850_1665226_ans_6ae5cb5dfa864d0d91aa5770f8e0edb7.png

A company manufactures two types of types of toys $A$ and $B$ . Type $A$ requires $5$ minutes each for cutting and $10$ minutes each for assembling. Type $B$ requires $8$ minutes each for cutting and $8$ hours minutes each for assembling. There are $3$ hours available for cutting and $4$ hours available for assembling in a day. The profit is $Rs. 50$ each on type $A$ and $Rs. 60$ each on type $B$ . How many toys of each type should the company manufacture in a day to maximize the profit?

1573852_1665233_ans_828aad6509d74a7ab1038cebc487099d.png

A dietician mixes together two kinds of food in such a way that the mixture contains at least $6$ units of vitamin $A,7$ units of vitamin $B, 11$ units of vitamin $C$ and $9$ units of vitamin $D$ . The vitamin contents of $1\ kg$ of food $X$ and $1\ kg$ of food $Y$ are given below:
Vitamin $A$ Vitamin $B$ Vitamin $C$ Vitamin $D$
Food $X$ $1$ $1$ $1$ $2$
Food $Y$ $2$ $1$ $3$ $1$
One $kg$ of food $X$ costs $Rs\ 5$ , whereas one kg of food $Y$ costs $Rs\ 8$ . Find the least cost of the mixture which will produce the desired diet.

Let

$x\ kg$ of food

$X$ and

$y\ kg$ of food

$Y$ are mixed to produce the desired diet. The

$LPP$ is

Minimize

$Z=5x+8y$

Subject to

$x+2y\ge 6$

$x+y\ge 7$

$x+3y\ge 11$

$2x+y\ge 9$

and,

$x,y\ge 0$

A manufacturer produces two types of steel trunks requires $3$ hours on machine $A$ and $B$ . For completing, the first types of the trunk requires $3$ hours on machine $A$ and $3$ hours on machine $B$ . Machine $A$ and $B$ can work at most for $18$ hours and $15$ hours per day respectively. He earns a profit if $Rs. 30$ and $Rs. 25$ per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?

1573753_1664909_ans_eb7c64f161e646178ba6f0a2a4934484.png

There are two types of fertilizers $'A'$ and $'B'$ . $'A'$ consists of $12\%$ nitrogen and $5\%$ phosphoric acid whereas $'B'$ consists of $4\%$ nitrogen and $5\%$ phosphoric acid. After testing the soil conditions, farmer finds that he needs at least $12$ kg of nitrogen and $12$ kg of phosphoric acis for his crops. If $'A'$ costs Rs. $10$ per kg and $'B'$ cost Rs. $8$ per kg, then graphically determine how much of each type of fertilizers should be used sot hat nutrient requirements are met at a minimum cost.

Given
Ref image

Let the requirement of fertilizer 'A' by farmer be 'x' kg and that of B be 'y' kg. Given that farmer requires atleast

$12 kg$ of Nitrogen &

$12 kg$ of phosphoric acid for his crops.

The in equation thus formed based on problem is

$\left(\dfrac{12}{100}\right) x + \left(\dfrac{4}{100} \right) y \ge 12 \Rightarrow 12x + 4y \ge 1200$

$\Rightarrow 3x + y \ge 300$ ___(1)

Also

$\dfrac{5x}{100} + \dfrac{5y}{100} \ge 12 \Rightarrow 5x + 5y \ge 1200$

$\Rightarrow x + y \ge 240$ ___(2)

$x \ge 0 ; y \ge 0$ ___(3)

Total cost of fertiliser

$z = 10x + 8y$

Solving the LPP

$3x + y = 300$

$x + y = 240$

$x = 0$

$y = 0$

Ref. image

The minimum value of z is obtained at

$C(30, 210)$

$\therefore$ Requirement of 'A' is

$30 kg$ & 'B' is

$210 kg$ .

1959575_1665556_ans_ed44496e13a04f5a9fcfcc7b5a94484b.png

A factory makes tennis rackets and cricket bats. A tennis racket takes $1.5$ hours of machine time and $3$ hours of craftsman's time in its making while a cricket bat takes $3$ hour of machine time and $1$ hour of craftsman's time. In a day, the factory has the availability of not more than $42$ hours of machine time and $24$ hours of craftsman's time. If the profit on a racket and on a bat is Rs. $20$ and Rs. $10$ respectively, find the maximum profit of the factory when it works a full capacity.

Let us assume the factory makes

$x$ units of tennis rackets and

$y$ units of cricket bats.

A tennis racket takes

$1.5$ hours of machine time and

$3$ hours of craftsman's time.

Total machine time for

$x$ tennis rackets

$=1.5x=\frac{3}{2}x$

Total craftsman's time for

$x$ tennis rackets

$=3x$

A cricket bat takes

$3$ hours of machine time and

$1$ hours of craftsman's time.

Total machine time for

$y$ tennis rackets

$=3y$

Total craftsman's time for

$y$ tennis rackets

$=y$

Profit on one racket and one bat is

$Rs. 20$ and

$Rs.10$ respectively.

Thus total profit incurred

$=20x+10y$

The factory has the availability of

$42$ hours of machine time and

$24$ hours craftsman's time atmost, in a day.

Hence the constraints are-

$\frac{3}{2}x+3y \leq 42 \Rightarrow 3x+6y \leq 84$

$. . . . . . . . . . . . (i)$

$3x+y \leq 24$

$. . . . . . . . . . . . (ii)$

$x \geq 0, y \geq 0$

$. . . . . . . . . . . . (iii)$

Max

$Z=20x+10y$

Plotting these inequations we get the following corner points from the graph-

$A(0,14)$ ;

$B(4,12)$ ;

$C(8,0)$

$A$ ,

$Z=20 \times 0 + 10 \times 14=140$

$B$ ,

$Z=20 \times 4 + 10 \times 12 = 80 +120 =200$

$C$ ,

$Z= 20 \times 8 + 10 \times 0 =160$

Thus,

$Z$ is maximum at

$B(4,12)$

Thus in a day, 4 units of tennis rackets and 12 units of cricket bats must be produced to gain maximum profit.

1620674_1665396_ans_50990b863ebd4441b3430f2a8bd0c8af.PNG

Find the maximum value of $Z = 7x + 7y$ , subject to the constraints $x\geq 0, y \geq, x + y \geq 2$ and $2x + 3y \leq 6$ .

1573032_1708383_ans_8a5e85b66d2c442f90c7a66b7ac028e5.png

Solve the following linear programming problem:
Maximise: $z=150x+250y$
Subject to: $4x+y \le 40$
$3x+2y\le 60$
$x\ge 0$
$y\ge 0$

Solve the following linear programming problem:

Maximise:

$z=150x+250y$

Subject to:

$4x+y \le 40$

$3x+2y\le 60$

$x\ge 0$

$y\ge 0$

Refer image.

$(0,30)\Rightarrow z=150\times 0+250\times 30=7500$

$(4,24)\Rightarrow z=150\times 4+250\times 24=6600$

$(10,0)\Rightarrow z=150\times 10+250\times 0=1500$

$\therefore max(z)=7500$ at

$(0,30)$

1497047_1700321_ans_8dfd9aac82454a698ab04ba3a822445e.png

Show that the solution set of the following linear constraints is empty:
$x - 2y \geq 0, 2x - y \leq -2, x \geq 0$ and $y\geq 0$ .

$C\text{onverting the inequations to equations we obtain}$

$\text{x-2y=0,2x-y=-2,x=0,y=0}$

$\text{x-2y=0}$

$\text{This line meet the x-axis at }\left( 0,0 \right)and\,\text{ y-axis}\,\text{at}\left( 0,0 \right).\,\text{ if x=1 then y=1/2}\text{. }$

$\text{si we have another point}\left( 1,1/2 \right).\,D\text{raw a thick line through }\left( 0,0 \right)and\left( 0,1/2 \right)$

$We\text{ see that the}\,\left( 1,0 \right)\,satisfies\,\text{the }\,\text{inequation}\,\text{x-2y}\ge \text{0}\,\text{.Hence portion containing }\left( 1,0 \right)\text{ represents}$

$\text{the solution set of the inequation}\,\text{x-2y}\ge \text{0}$

$The\text{ line 2x-y=-2}\,\text{ meets the x-axis}\,\text{at }\left( -1,0 \right)\,\text{ and y-axis}\,\text{at}\left( 0,2 \right).\,Dr\text{aw a thick line joining}\left( -1,0 \right)and\left( 0,2 \right).$

$we\text{ see that the origin}\,\left( 0,0 \right)\text{ does not satisfy the inequation}\,2x-y\le -2$

$clearly\text{ x}\ge 0\,\text{and }\,\text{y}\ge 0\text{ represents the first quadrant}$

$Hence\text{ there is no common region in all the lines}\text{.Hence the solution set to given set of }$

$\text{inequations is empty}\text{.}$

1548436_1708373_ans_02aa98f9f4cf4c0f8dd1075efa04d67d.JPG

A company manufactures two articles $A$ and $B$ . There are two department through which these article are processed : (i) assembly and (ii) finishing department. The maximum capacity of the first department is $60$ hours a week and that of other department. $48$ hours, per week. The product of each unit of article $A$ requires $4$ hours in assembly and $2$ hours in finishing and that of each unit of $B$ requires $2$ hours in assembly and $4$ hours in finishing. If the profit is Rs. $6$ for each unit $A$ and Rs. $8$ for each unit of $A$ and Rs. $8$ for each unit of $B$ , find the number of units of $A$ and $B$ to be produced per week in order to have maximum profit.

Let the number of type

$A$ and

$B$ articles produced be

$x$ and

$y$ respectively.

Each unit of article

$A$ requires

$4$ hours in assembly and

$2$ hours in the finishing department.

Total time taken by

$x$ units of article

$A$ in assembly department

$= 4x$

Total time taken by

$x$ units of article

$A$ in finishing department

$= 2x$

Each unit of article

$B$ requires

$2$ hours in assembly and

$4$ hours in the finishing department.

Total time taken by

$y$ units of article

$B$ in assembly department

$= 2y$

Total time taken by

$y$ units of article

$B$ in finishing department

$= 4y$

Maximum capacity of assembly department =

$60$ hours/week

Maximum capacity of finishing department =

$48$ hours/week

Profit on article

$A$ =

$Rs.6/$ unit

Profit on article

$B$ =

$Rs.8/$ unit

Total profit in a week =

$6x+8y$

Thus the required constraints are-

$4x+2y \leq 60$

$. . . . . . . . . . . . . (i)$

$2x + 4y \leq 48$

$. . . . . . . . . . . . . (ii)$

$x \geq 0, y \geq 0$

$. . . . . . . . . . . . . (iii)$

Max

$Z=6x+8y$

Plotting these inequations on the graph we get the following corner points-

$A(0,12)$ ;

$B(12,6)$ ;

$C(15,0)$

$A$ ,

$Z=6 \times 0 + 8 \times 12=96$

$B$ ,

$Z= 6 \times 12 + 8 \times 6 = 72 + 48 =120$

$C$ ,

$Z= 6 \times 15 + 8 \times 0 =90$

Thus,

$Z$ is maximum at

$B(12,6)$

Thus, 12 units of article

$A$ and 6 units of article

$B$ should be produced to get the maximum profit per week.

1621833_1665239_ans_a0c37eed65e2480cad86bd3fdae4ee68.PNG

A medical company has factories at two places, $A$ and $B$ . From these places, supply is made to each of its three agencies situated at $P,Q$ and $R$ . The monthly requirement of the agencies are respectively $40, 40$ and $50$ packets of the medicines, while the production capacity of the factories, $A$ and $B$ are $60$ and $70$ packets respectively. The transportation cost per packet from the factories to the agencies are given below:
Transportation cost per packet (in Rs)
To $\downarrow$ \ From $\to$ $A$ $B$
$P$ $5$ $4$
$Q$ $4$ $2$
$R$ $3$ $5$
How many packets from each factory be transported to each agency so that the cost of transportation is minimum cost?

1573860_1665582_ans_e73a73ebac0e43d2a1aff27f30c23c25.png

Find the linear constraints for which the shaded area in the figure given is the solution set.

1573031_1708377_ans_a7d3d2964b094f6eb8ba3615c8c5f1b9.png

A man owns a field of area $1000\ m^{2}$ . He wants to plant fruit trees in it. He has a sum of $Rs. 1400$ to purchase young trees. He has the choice of two types of trees. Type $A$ requires $10\ m^{2}$ of ground per tree and costs $Rs. 20$ per tree, and type $B$ requires $20\ m^{2}$ of ground per tree and costs $Rs. 25$ per tree. When full grown, a type- $A$ tree produces and average of $20\ kg$ of fruit which can be sold at a profit of $Rs. 2$ per kg and a type - $B$ tree produces an average of $40\ kg$ of fruit which can be sold at a profit of $Rs. 1.50$ per kg. How many of each type should be planted to achieve maximum profit when trees are fully grown? What is the maximum profit?

1573761_1708508_ans_ef12803dbd0e4b54acdb18224215023e.png

A manufacturer produces two types of soap bars using two machines, $A$ and $B$ . $A$ is operated for $2$ minutes and $B$ for $3$ minutes to manufacture the first type, while it takes $3$ minutes on machine $A$ and $5$ minutes on machine $B$ to manufacture the second type. Each machine can be operated at the most for $8$ hours per day. The two types of soap bars are sold at a profit of $Rs. 0.25$ and $Rs. 0.50$ each. Assuming that the manufacturer can sell all the soap bars he can manufacture, how many bars of soap of each type should be manufactured per day so as to maximize his profit?

1573682_1708457_ans_9ce4218127564de89982d5e2e0e6e929.png

Maximize $Z = 3x + 5y$ , subject to the constraints $x + 2y \leq 2000, x + y \leq 1500, y \leq 600, x \geq 0$ and $y\geq 0$ .

1573063_1708403_ans_806259f364ab436c9b8c858db384e9dd.png

Mr Dass wants to invest $Rs. 12000$ in Public Provident Fund (PPF) and in National bonds. He has to invest at least $Rs. 1000$ in PPF and at least $Rs. 2000$ in bonds. If the rate of interest on PPF is $12\%$ per annum and that on bonds is $15\%$ per annum, how should he invest the money to earn maximum annual income? Also find maximum annual income.

1573070_1708424_ans_ec14350f232b42b0b82bf09dfcf382cc.png

Maximize $Z = 4x + 9y$ , subject to the constraints $x\geq 0, y \geq 0, x + 4y \leq 200, 2x + 3y \leq 134$ .

1573034_1708387_ans_1e5d07d85b5d4949b3d8825abc94dd59.png

Minimize $Z = 2x + 3y$ , subject to the constraints $x \geq 0, y \geq 0, x + 2y \geq 1$ and $x + 2y \leq 10$ .

1573057_1708396_ans_d7b935598ac2425da35bcf6b83a194df.png

A company producing soft drinks has a contract which requires a minimum of $80$ units of chemical $A$ and $60$ units of chemical $B$ to go in each bottle of the drink. The chemicals are available in a prepared mix from two different suppliers. Supplier $X$ has a mix of $4$ units of $A$ and $2$ units of $B$ that costs $Rs. 10$ , and the supplier $Y$ has a mix of $1$ unit of $A$ and $1$ unit of $B$ that costs $Rs. 4$ . How many mixes from $X$ and $Y$ should the company purchase to honour the contract requirement and yet minimize the cost?

1574202_1708461_ans_be27a0202a52499b8af3797bd38f0111.png

A firm manufactures two types of products, $A$ and $B$ , and sells them at a profit of $Rs. 2$ on type $A$ and $Rs. 2$ on type $B$ . Each product is processed on two machines, $M_{1}$ and $M_{2}$ . Type $A$ requires one minute of processing time on $M_{1}$ and two minutes on $M_{2}$ . Type $B$ requires one minute on $M_{1}$ and one minute on $M_{2}$ is available for at most $10$ hours a day.
Find how many producs of each type the firm should produce each day in order to get maximum profit.

1573672_1708456_ans_103063debc014671b043ff6ef8928c7d.png

Find the minimum value of $Z = 3x + 5y$ , subject to the constraints $-2x + y \leq 4, x + y \geq 3, x - 2y \leq 2, x \geq 0$ and $y\geq 0$ .

2016108_1708390_ans_ebd980ec801744788cf192c9952479b3.png

Find the maximum and minimum values of $Z = 2x + y$ . subject to the constraints
$x + 3y \geq 6, x - 3y \leq 3, 3x + 4y \leq 24, -3x + 2y \leq 6, 5x + y \geq 5, x \geq 0$ and $y \geq 0$ .

1573067_1708414_ans_f4244c32bbb941c2a25260ed84f43de0.png

An aeroplane can carry a maximum of $200$ passengers. A profit of $500 \ Rs.$ is made on each executive class ticket out of which $20 \%$ will go to the welfare fund of the employees. Similarly a profit of $400 \ Rs.$ is made on each economy ticket out of which $25 \%$ will go for the improvement of facilities provided to economy class passengers. In both cases, the remaining profit goes to the airline's fund. The airline reserves at least 20 seats for executive class. However, at least four times as many passengers prefer to travel by economy class than by the executive class. Determine, how many tickets of each type must be sold in order to maximise the net profit of the airline. Make the above as an LPP and solve graphically.

Let there be

$x$ tickets of executive class and

$y$ tickets of economy class be sold. Let

$Z$ be net profit of the airline.

Here, we have to maximise

$Z$ .

Now,

$Z= 500z \times \frac{80}{100} +400y \times \frac{75}{100}$

$Z=400x + 300y$

$...(i)$

Subject to constraints:

$x \geq 20$

$...(ii)$

Also

$x+y \leq 200$

$...(iii)$

$x + 4x \leq 200$

$[\therefore y = 4x]$

$\implies \ 5x \leq 200$

$\implies x \leq 40$

$...(iv)$

Shaded region is feasible region having corner points

$A(20,0), \ B(40,0), \ C(40,160), \ D(20, 180), \ E(0, 200)$ .

Now, value of

$Z$ is calculated at corner point as in the following table.

Hence, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the net profit of the airlines.

1640564_1783292_ans_162990cdcd9d467aab98531675a74e59.png

1640564_1783292_ans_162990cdcd9d467aab98531675a74e59.png

A brick manufacturer has two depots, $P$ and $Q$ , with stocks of $30000$ and $20000$ bricks respectively. He receives order from three buildings $A, B, C$ for $15000, 20000$ and $15000$ bricks respectively. The cost of transporting $1000$ bricks to the buildings from the depots are given below.
Cost of transportation
(in Rs. per $1000$ bricks)
From To A B C
$P$ $40$ $20$ $30$
$Q$ $20$ $60$ $40$
How should the manufacturer fulfil the orders so as to keep the cost of transportation minimum?

1573906_1708532_ans_b3aea4cedc7c4120a21436a7a9389f7a.png

Maximise $z = 8x + 9y$ subject to the constraints given below :
$2x + 3y \leq 6; 3x - 2y \leq 6; y \leq 1; x\geq 0; y \geq 0$

Given constraints are

$2x + 3y \leq 6$

$3x - 2y \leq 6$

$y \leq 1$

$x, y \geq 0$

For graph of

$2x +3y \leq 6$
We draw the graph of

$2x + 3y =6$

$x$

$0$

$3$

$y$

$2$

$0$

Putting Origin in the equation, we get:

$2 \times 0 + 3 \times 0 \leq 6 \implies (0,0)$ satisfy the constraints.
Hence, feasible region lie towards origin side of line.

For graph of

$3x-2y \leq 6$
We draw the graph of line

$3x-2y=6$ .

$x$

$0$

$2$

$y$

$-3$

$0$

Putting Origin in the equation, we get:

$3 \times 0 - 2 \times 0 \leq 6$

$\implies$ Origin

$(0,0)$ satisfy

$3x-2y =6$ .
Hence, feasible region lie towards origin side of line.

For graph of

$y \leq 1$
We draw the graph of line

$y=1$ , which is parallel to

$x-$ axis and meet

$y-$ axis at

$1$ .

Putting Origin in the equation, we get:

$0 \leq 1 \implies$ feasible region lie towards origin side of

$y=1$ .

Also,

$x \geq 0, y \geq 0$ says feasible region is in Ist quadrant
Therefore,

$OABCDO$ is the required feasible region, having corner point

$O(0,0), \ A(0,1), \ B \left(\dfrac{3}{2}, 1\right), \ C\left(\dfrac{30}{13}, \dfrac{6}{13}\right), \ D(2,0)$ .

Here, feasible region is bounded. Now the value of objective function

$Z=8x+9y$ is obtained as.
Corner Points

$Z=8x+9y$

$O(0,0)$

$0$

$A(0,1)$

$9$

$B \left(\dfrac{3}{2}, 1\right)$

$21$

$C \left(\dfrac{30}{13}, \dfrac{6}{13}\right)$

$22.6$

$D(2,0)$

$16$

$Z$ is maximum when

$x=\frac{30}{13}$ and

$y= \frac{6}{13}$ .

1638715_1782715_ans_3b98ae26ded44385ac9ec28b7c07fd9f.jpg

A manufacturer has $640$ litres of a $8\%$ solution of boric acid. How many litres of a $2\%$ boric acid solution be added to it so that the boric acid content in the resulting mixture will be more than $4\%$ but less than $6\%$ ?

Let

$x$ litres of

$2\%$ boric acid added to

$640$ litres of

$5\%$ solution of boric acid.

Then, total mixture

$=(640+x)$ litres

Boric acis content in this mixture

$=\left(\dfrac{640\times 8}{100}+\dfrac{x\times 2}{100}\right)$ litres.

$\therefore \dfrac{4}{100}\times (640+x)<\left\{\dfrac{640\times 8}{100}+\dfrac{x\times 2}{100}\right\}<\dfrac{6}{100}\times (640+x)$

$\Rightarrow 2560+4x<5120+2x<3840+6x$

$\Rightarrow 2x<2560$ and

$4x>1280\Rightarrow 320<x<1280$ .

A company manufactures cassettes. Its cost and revenue functions are $C(x)=26000+30x$ and $R(x)=43x$ respectively, where $x$ is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit?

2016152_1709435_ans_c4baec4f05414a31b47a5d866b3ea425.png

How many litres of water will have to be added to $600$ litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ , but less than $30\%$ acid content?

Let

$x$ litres of water may be added.

Then,

$25\%$ of

$(600+x)<\dfrac{600\times 45}{100}<30\%$ of

$(600+x)$

$\Rightarrow \dfrac{600+x}{4}<270<\dfrac{1800+3x}{10}$

$\Rightarrow \dfrac{600+x}{4}<270$ and

$\dfrac{1800+3x}{10}>270$

$\Rightarrow 600+x<1080$ and

$1800+3x>2700$

$\Rightarrow 300 < x<480$ .

The water acidity in a pool is considered normal when the average $pH$ reading of three daily measurements between $8.2$ and $8.5$ . If the first two $pH$ readings are $8.48$ and $8.35$ , find the range of the $pH$ values for the third reading that will result in the acidity level being normal.

Let the third reading be

$x$ .

Then,

$8.2<\dfrac{8.48+8.35+x}{3}<8.5$

$\Rightarrow 24.6<16.83+x<25.5$

$\Rightarrow 24.6-16.83<x<25.5-16.83$

$\Rightarrow 7.77<x<8.67$ .

A village has 500 hectare of land to grow two types of plants, $X$ and $Y$ . The contributions of total amount of oxygen produced by plant $X$ and plant $Y$ are $60 \%$ and $40 \%$ per hectare respectively. To control weeds, a liquid herbicide has to be used for $X$ and $Y$ at rates of $20$ litres and $10$ litres per hectare, respectively. Further no more than $8000$ litres of herbicides should be used in order to protect aquatic animals in a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total production of oxygen?

Let

$x$ and

$Y$ be grown in

$x$ and

$y$ hectares.

So,

$x \geq 0$ and

$y \geq 0$

$x+y \leq 500$

$...(i)$

Contribution of oxygen by the plants

$=60 \%$ of

$x+ 40 \%$ of

$y$

$Z= \frac{6x}{10} + \frac{4y}{10} = 0.6x+0.4y$

Also, Amount of liquid herbicides required

$=(20x +10 y)$ litres

Given

$20x+10y \leq 800$

$...(ii)$

The LPP for given problem is
Maximum,

$Z=0.6x+0.4y$
Subject to constraints:

$x+y \leq 500$

$...(iii)$

and

$2x +y \leq 800$

$...(iv)$

$x,y \geq 0$

Sketching a graph for the above LPP, we get the region shown in the figure.
Solving

$x+y =500$ and

$2x+y=800$ , we get

$x=300$ and

$y=200$ as

$B(300,200)$

Maximum production of oxygen will be achieved when plant

$X$ is planted in

$300$ hectares and plant

$Y$ is planted in

$200$ hectares.
1640570_1783280_ans_7fdd83fdfdf040a591d0ae588dc15fd6.png

A firm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. They need certain nutrients, names as $X, Y, Z$ . The pigs are fed on two products, $A$ and $B$ . One unit of product $A$ contains $36$ units of $X, 3$ units of $Y$ and $20$ units of $Z$ , while one unit of product $B$ contains $6$ units of $X, 12$ units of $Y$ and $10$ units of $Z$ . The minimum requirements of $X, Y, Z$ are $108\ units, 36\ units$ and $100\ units$ . respectively. Product $A$ costs $Rs. 20$ per unit and product $B$ costs $Rs. 40$ per unit. How many uits of each product must be taken to minimize the cost? Also, find the minimum cost.

1574220_1708558_ans_2f957d37d1b741d293cd17608708eee0.png

Determine the maximum value of $Z = 3x + 4y$ if the feasible region (shaded) for a LPP is shown in Fig.

$\begin{matrix} \text{Corner Points}& \text{Corresponding value of Z} \\ (0,0) &0 \\(52, 0) &156 \\ (44, 16) & 196 \ (Maximum) \\ (0, 38) & 152 \end{matrix}$

As clear from the graph, corner points are O, A, E and D with coordinate (0, 0), (52, 0), (144, 16) and (0, 38), respectively. Also given region is bounded.

Here,

$Z = 3x + 4y$

$\therefore 2x + y = 104$ and

$2x +4y = 152$

solving above equations, we get

$\Rightarrow -3y = - 48$

$\Rightarrow y = 16$ and

$x = 44$

Hence, Z is maximum at (44, 16) and its maximum value is 196.

The feasible region for a LPP is shown in Fig. 12.Find the minimum value of $Z = 11x + 7y$ .

From the figure, it is clear that feasible region is bounded with coordinate of corner points as (0,3), (3, 2) and (0, 5). Here

$Z = 11x + 7y$

$\therefore x + 3y = 9$ and

$x + y = 5$

solving above equations, we get

$\Rightarrow 2y = 4$

$\therefore y = 2$ and

$x = 3$

So, intersection points of

$x + y = 5$ and

$x + 3y = 9$ is (3, 2).

Hence, the minimum value of Z is 21 at (0, 3).

1906481_1799058_ans_51f9cef10d1a483f897eae48b2f69d1c.png

Feasible region (shaded) for a LPP is shown in Fig., Maximise $Z = 5x + 7y$ .

The Shaded region is bounded and has coordinate of corner points as

(0, 0), (7,0), (3, 4) and (0, 2), Also

$Z = 5x + 7y$ .

Hence, the maximum value of Z is 43 at (3, 4)

1904729_1799036_ans_5d90b47cb8944fcfaceb333b390c8342.PNG

Maximise $Z = 3x + 4y$ , subject to the constraints : $x + y \leq 1, x \geq 0, y \geq 0$ .

Maximise

$Z = 3x + 4y$ , subject to the constraints

$x + y \leq 1, x \geq 0, y \geq 0$ .

The Shaded region shown in figure as OAB is bounded and the coordinates of corner points O, A and B are (0, 0), (1, 0) and (0, 1) respectively.

Hence, the maximum value of Z is 4 at (0, 1)

1904745_1798995_ans_d5905cdf74804b5db5ab0cedc6346a88.PNG

Minimise $Z = 13x - 15y$ subject to the constraints :
$x + y \leq 7, 2x - 3y + 6 \geq 0, x \geq 0, y \geq 0$ .

Minimise

$Z = 13x - 15y$ subject to the constraints :

$x + y \leq 7, 2x - 3y + 6 \geq 0, x \geq 0, y \geq 0$ .

Replace (0, 7) by (7, 0) in horizontal line.

shaded region shown as OABC is bounded and coordinates of its corner points are (0, 0), (7, 0), (3, 4) and (0, 2) respectively.

Corner Points Corresponding value of

$Z$

(0, 0) 0

(7, 0) 91

(3, 4) -21

(0, 2) -30 (Minimum)

1904738_1799022_ans_43906ae531ff4ac6bb421775d73aa694.PNG

Maximise the function $Z = 11x + 7y$ , subject to the constraints:
$x \leq 3, y \leq 2, x \geq 0, y \geq 0$

Maximise

$Z = 11x + 7y$ , subject to the constraints

$x \leq 3, y \leq 2, x \geq 0, y \geq 0$

The shaded region as shown in the figure as OABC is bounded and the coordinates of corner points are (0, 0), (3, 0), (3, 2) and (0, 2), respectively.

Hence, Z is maximise at (3, 2) and its maximum value is 47.

1904743_1799013_ans_166d423c52984bd5964c63b0fc4fda53.PNG

The feasible region for a LPP is shown in Fig. Evaluate $Z = 4x + y$ at each of the corner points of this region. Find the minimum value of Z, if it exists.

$\begin{matrix} \text{Corner Points}&\text{Corresponding value of Z} \\ (4,0) &16 \\ (2,1) & 9 \\ (0,3) & 3\ (Minimum) \end{matrix}$

From the shaded region, it is clear that feasible region is unbounded with the corner points A(4, 0, B(2, 1) and C(0, 3).

Also, we have

$Z = 4x + y$ ,

Now, we see that 3 is the smallest value of Z at the corner point (0, 3). Note that here we see that the region is unbounded, therefore 3 may or may not be the minimum value of Z.

To decide this issue, we graph the inequality

$4x + y < 3$ and check whether the resulting open half plan has no point in common with feasible region otherwise, Z has no minimum value.

From the shown graph above, it is clear that there is no point in common with feasible region and hence Z is minimum value of 3 at (0, 3).

1905193_1799064_ans_77d4ec8d105b409cad305ee4435ee632.PNG

A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is Rs. 50 and that on type B circuit is Rs. 60, formulate this problem as a LPP so that the manufacturer can maximise his profit.

Let the manufacturer produce x units of type A circuit and y units of type B circuits.

From the given information, we have following corresponding constraint table.

Thus, we see that total profit

$Z= 50x + 60y$ (in Rs.)

Now, we have the following mathematical model for the given problem.

Maximise

$Z = 50x + 60y$ ...........(i)

Subject to the constraints.

$20x + 10y \leq 200$ [resistor constraint]

$\Rightarrow 2x + y \leq 20$ ......(ii)

And

$10x + 20y \leq 120$ [transistor constraint]

$\Rightarrow x + 2y \leq 12$ .......(iii)

And

$10x + 30y \leq 150$ [capacitor constraint]

$\Rightarrow x + 3y \leq 15$ ..... (iv)

And

$x \leq 0, y \leq 0$ .... (v),

$x \geq 0, y \geq 0$ .... (v) [non[negative constraint]

So, Maximise

$Z = 50x + 60y$ , subject to

$2x + y \leq 20, x + 2y \leq 12, x + 3y \leq 15, x \leq 0, y \leq 0, x \geq 0, y \geq 0$

Determine the maximum value of $Z = 11x + 7y$ subject to the constraints : $2x + y \leq 6, x \leq 2, x \geq 0, y \geq 0$ .

We have, maximise

$Z = 11x + 7y$ ..........(i)

Subject to constraints

$2x + y \leq 6$ .........(ii)

$x \leq 2$ ........ (iii)

$x \geq 0, y \geq 0$ .........(iv)

Refer table and image

We see that, the feasible region as shaded determined by the system of constraint (ii) to (iv) is OABC and is bounded. So, now we shall use corner point method to determine the maximum value of z.

Hence, the maximum value of Z is 42 at (0, 6).

1904747_1798991_ans_2d4058b6c2c246baa7a6a8351f5765e3.PNG

In Fig., the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of $Z = x + 2y$ .

$\begin{matrix} \text{Corner Point}&\text{Corresponding value of Z} \\ \left(\dfrac{3}{13}, \dfrac{24}{13}\right)& \dfrac{3}{13} + \dfrac{48}{13} = \dfrac{51}{13} = 3\dfrac{12}{13} \\ \left(\dfrac{18}{7}, \dfrac{2}{7}\right) & \dfrac{18}{7} + \dfrac{4}{7} = \dfrac{22}{7} = 3 \dfrac{1}{7} (Minimum) \\ \left(\dfrac{7}{2}, \dfrac{3}{4}\right) & \dfrac{7}{2} + \dfrac{6}{4} = \dfrac{20}{4} = 5 \\ \left(\dfrac{3}{2}, \dfrac{15}{4}\right) & \dfrac{3}{2} + \dfrac{30}{4} = \dfrac{36}{4} = 9 (Maximum)\end{matrix}$

From the shaded bounded region, it is clear that the coordinates of corner points are

$\displaystyle \left ( \frac{3} {13}, \frac{24} {13} \right ), \left ( \frac{18} {7}, \frac{2} {7} \right ), \left ( \frac{7} {2}, \frac{3} {4} \right )$ and

$\displaystyle \left ( \frac{3} {2}, \frac{15} {4} \right )$

Also, we have to determine maximum and minimum value of

$Z = x + 2y$ .

Hence, the maximum and minimum value of are 9 and

$3 \dfrac{1} {7}$ respectively.

A company manufactures two types of screw A and B. All the screws have to pass through a threading machine and a slotting machine. A box type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. A box of type B screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours.
On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws.
Formulate this problem as a LPP given that the objective is to maximise profit.

Let the company manufacture x boxes of type A screws and y boxes of type screws, From the gives information, we have following corresponding constraint table

Thus, we see that objective function for maximum profit is

$Z = 100x + 170y$ .

Subject to constraints.

$$2x + 8y \leq 60 \times 60 [time constraint for threading machine]

$\Rightarrow x + 4y \leq 1800$ .......... (i)

$$3x + 2y \leq 60 \times 60 [time constraint for slotting machine]

$\Rightarrow 3x + 2y \leq 3600$ .......... (ii)

Also,

$x \leq 0, y \leq 0, x \geq 0, y \geq 0$ [non-negative constraint ] .... (iii)

$\therefore$ required LPP is,

Maximise

$Z = 100x + 170y$

Subject to constraints

$x + 4y \leq 1800, 3x + 2y \leq 3600, x \leq 0, y \leq 0$

$x + 4y \leq 1800, 3x + 2y \leq 3600, x \geq 0, y \geq 0$

1906484_1799207_ans_cb24239bc673446da5741d6bc6321f26.png

Fill in the blank.
If the feasible region for a LPP is _______, then the optimal value of the objective function Z = ax + by may or may not exist.

If the feasible region for a LPP is is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

Fill in the blank.
In a LPP, the objective function is always _________ .

In a LPP, the objective function is always linear.

A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the pertrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol one hour time. He wishes to find the maximum distance that he can travel.
Express this problem as a linear programming problem.

Let the man rides to his motorcycle to a distance x km at the speed of 50 km/h and to a distance y km at the speed of 80 km/h.

Therefore, cost on petrol is

$2x + 3y$ .

Since, he has to spend Rs 120 at most on petrol.

$\therefore 2x + 3y \leq 120$ .... (i)

Also, he has at most 1 hour time.

$\therefore \frac{x} {50} + \frac{y} {80} \leq 1$

$\Rightarrow 8x + 5y \leq 400$ .....(ii)

Also, we have

$x \geq 0, y \geq 0$ [non-negative constrants]

Thus, required LPP to travel maximum distance by him is

Maximise

$Z = x + y$ , subject to

$2x + 3y \leq 120, 8x + 5y \leq 400, x \geq 0, y \geq 0$

A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more thanThe company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit of the company.

Let the company manufactures x number of type A sweaters and y number of type B sweaters.

From the given information we see that cost to make a type A sweater is Rs 360 and cost to make a type B sweater is Rs 120.

Also, the company spent at most Rs 72000 a day

$\therefore 360x + 120y \leq 72000$

$\Rightarrow 3x + y \leq 600$ ......(i)

Also, company can make at most 300 sweaters.

$\therefore x + y \leq 300$ .....(ii)

Further, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100 i.e.

$x + 100 \geq y$

$\therefore x - y \geq -100$ .... (iii)

Also, we have non-negative constraints for x and y i.e.

$x \geq 0, y \geq 0$ ........(iv)

Hence, the company makes a profit of Rs 200 for each sweater of type A and Rs 120 for each sweater of type B i.e.

Profit (Z) =

$200x + 120y$

Thus, the required LPP to maximise the profit is Maximise

$Z = 200x + 120y$ is subject to constraints.

$3x + y \leq 600$

$x = y \leq 300$

$x - y \geq -100$

$x \geq 0, y \geq 0$

Fill in the blank.
A corner point of a feasible region is a point in the region which is the _________ of two boundary lines.

A corner point of a feasible region is a point in the region which is the intersection of two boundary lines.

Refer to ExerciseHow many of circuits of Type A and Type B, should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

$\begin{matrix} \text{Corner Ponts} & \text{Corresponding value of }Z = 50x + 60y \\ (0, 0) & 0 \\ (10, 0) & 500 \\ \left( \dfrac{28}{3}, \dfrac{4}{3}\right) & \dfrac{1400}{3} + \dfrac{240}{3} = \dfrac{1640}{3}=546.66(maximum) \\ (6, 3) & 480 \\ (0, 5) & 300 \end{matrix}$

Referring to solution 11, we have to

Maximise

$Z= 50x + 60y$ , subject to

$2x + y \leq 20, x + 2y \leq 12, x + 3y \leq 15, x \leq 0, y \leq 0, x \geq 0, y \geq 0$

From the shaded region it is clear that the feasible region determined by the system of constraint is OABCD and it is bounded and the coordinates of corner points are

(0, 0), (10, 0),

$\left ( \frac{28} {3}, \frac{4} {3} \right)$ , (6, 3) and (0, 5), respectively.

[Solving x + 2y = 12 and 2x + y = 20

$\Rightarrow x = \frac{28} {3}, y = \frac{4} {3}$ and x + 3y = 15 and x + 2y = 12

$\Rightarrow$ y = 3 x = 16]

Since, the manufacturer is required to produce two types of circuits A and B and it is clear that parts of resistor, transistor and capacitor cannot be in fraction, So the required maximum profit is 480 where circuits of type A is 6 and circuits of type B is 3.

1906458_1799797_ans_e1ca94f0cdcb4b20b406230fb3142771.png

Fill in the blank.
A feasible region of a system of linear inequalities is said to be ________ if it can be enclosed within a circle.

A feasible region of a system of linear inequalities is said to be bounded, if it can be enclosed within a circle.

Fill in the blank.
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same ________ value.

In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same maximum value.

Fill in the blank.
The feasible region for an LPP is always a _________ polygon.

The feasible region for an LPP is always a convex polygon.

Refer to ExerciseWhat will be the minimum cost?

$\begin{matrix} \text{Corner Points} & \text{Corresponding value of} \ X = 400x + 200y\\ (0, 15) &3000 \\ (5,5) & 3000 \\ \left(\dfrac{30}{7}, \dfrac{30}{7}\right) & 400\times \dfrac{30}{7} + 200 \times \times \dfrac{30}{7} = \dfrac{18000}{7} \\ & = 2571.43\ (minimum)\end{matrix}$

Referring to solution 12, we have minimise Z = 400x + 200y, subject to

$5x + 2y \geq 30$ .

$2x + y \geq 15, x \leq y, x \leq 0, y \leq 0$ ,

On solving x - y = 0 and 5x + 2y = 30, we get

$y = \frac{30} {7}, x = \frac{30} {7}$

On solving x - y = 0 and 2x + y = 15, we get x = 5, y = 5

So, from the shaded feasible region it is clear that coordinates of corner points are (0, 15), (5, 5) and

$\left (\frac{30} {7}, \frac{30} {7} \right)$

Hence, the minimum cost is Rs 2571.43.

1906461_1799804_ans_8021d55dbd7d49aeaf9726e6885a2ff0.png

A company makes 3 model of calculators : A, B and C at factory I and factory II. The company has ordered for at least 6400 calculators of model A, 4000 calculators of model B and 4800 calculators of model C. At factory I, 50 calculators of model A, 50 calculators of model B and 30 calculators of model C are made every day; at Factory II, 40 calculators of model A, 20 calculators of model B and 40 calculators of model C are made everyday. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.

$\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{value of} \ Z = 12000x + 15000y} \\ (160, 0) & 160\times 12000 = 1920000 \\ (80, 60) & (80\times 12 + 60 \times 15)\times 1000 = 1860000 \ (Minimum) \\ (32, 120) & (32\times 12 + 120 \times 15) \times 1000 = 2184000 \\ (0, 200) & 0 + 200 \times 15000 = 3000000\end{matrix}$

Let the factory I operate for x days and the factory II operate for y days.

At factory I, 50 calculators of model A and at the factory II, 40 calculators of model A are made everyday. Also, company has ordered for at least 6400 calculators of model A.

$50x + 40y \geq 6400 \Rightarrow 5x + 4y \geq 640$ ...... (i)

Also, at factory I, 50 calculators of model B and at factory II, 20 calculators of model B are made everyday.

Since, the company has ordered at least 4000 calculators of model B.

$50x + 20y \geq 4000 \Rightarrow 5x + 4y \geq 400$ ...... (i)

Similarly, for model C,

$30x + 40y \geq 4800 \Rightarrow 3x + 4y \geq 480$ ...... (iii)

[Since, x and y are non-negative]

It costs Rs 12000 and Rs 15000 each day to operate factories I and II, respectively.

$\therefore$ Corresponding LPP is

Minimise Z = 12000x + 15000y, subject to

$5x + 4y \geq 640$

$5x + 2y \geq 400$

$3x + 4y \geq 480$

$x \geq 0, y \geq 0$

On solving 3x + 4y = 480 and 5x + 4y = 640, we get x = 80 and y = 60.

On solving 5x + 4y = 640 and 5x + 2y = 400, we get x = 32 and y = 120.

Thus, from the graph, it is clear that feasible region is unbounded and the coordinates of corner points A, B, C and D are (160, 0), (80, 60), (32, 120) and (0, 200), respectively.

From the above table, it is clear that for given unbounded region the minimum value of Z may or may not be 1860000.

Now, for deciding this, we graph the inequality

12000x + 15000y < 1860000

$\Rightarrow 4x + 5y < 620$

And check whether the resulting open half plane has point in common with feasible region or not.

Thus, as shown in the figure, it has no common points so, Z = 12000x + 15000y has minimum value of 1860000.

So, number of days for factory I should be operated is 80 and number of days for factory II should be operated is 60 for minimum cost and satisfying the given constraints.

1905039_1799909_ans_ec69bb2677d84a26aaaef7aa8cdc554e.PNG

Refer to ExerciseHow many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

$\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{Corresponding value of} \ Z = 200x + 120y} \\ (0, 0) & 0 \\ (200, 0) & 40000 \\ (150, 150) & 150 \times 200 + 120 \times 150 = 48000\ (maximum) \\ (100, 200) & 100 \times 200 + 120 \times 200 = 44000 \\ (0, 100) & 120 \times 100 = 12000 \end{matrix}$

Referring to solution 14, we have to maximise Z = 200x + 120 y
Subject to

$x + y \leq 300, 3x + y \leq 600, x - y \geq -100, x \geq 0, y \geq 0$ .
On solving x + y = 300 and 3x + y = 600, we get
x = 150, y = 150
On solving x - y = -100 and x + y = 300, we get
x = 100, y = 200
From the shaded feasible region it is clear that coordinate of corner points are
(0, 0), (200, 0), (150,150), (100, 200) and (0, 100).
Hence, 150 sweaters of each type made by company and maximum profit is Rs 48000
1906425_1799817_ans_a53e6b7cba064e168897ce8c87f673e6.png

1906425_1799817_ans_a53e6b7cba064e168897ce8c87f673e6.png

Tablets Iron Calcium Vitamin
X
Y 6
2 3
3 2
4
In order to supplement daily diet, a person wishes to take some X and some wishes Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligram per tablet) are given as below.

The person needs at least 18 milligram of iron, 21 milligram of calcium and 16 milligram of vitamins. The price of each tablet of X and Y is Rs 2 and Re 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?

$\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{value of} \ Z = 2x + y} \\ (8, 0) & 16 \\ (6, 1) & 13 \\ (1, 6) & 8\ (Minimum) \\ (1, 9) & 9 \end{matrix}$

Let the person takes x units of tablet X and y unit of tablet Y.

So, from the given information, we have

$6x + 2y \geq 18 \Rightarrow 3x + y \geq 9$ ...... (i)

$3x + 3y \geq 21 \Rightarrow x + y \geq 7$ ...... (ii)

And

$2x + 4y \geq 16 \Rightarrow x + 2y \geq 8$ ...... (iii)

Also, we know that here,

$x \geq 0, y \geq 0$ .....(iv)

The price of each tablet of X and Y is Rs 2 and Rs 1, respectively,

So, the corresponding LPP is minimise Z = 2x + y, subject to

$3x + y \geq 9, x + y \geq 7, x + 2y \geq 8, x \geq 0, y \geq 0$

From the shaded graph, we see that for the shown unbounded region, we have coordinates of corner points A, B, C and D as (8, 0), (6, 1), (1, 6) and (0, 9), respectively.

[On solving x + 2y = 8 and x + y = 7, we get x = 6, y = 1 and on solving 3x + y = 9 and x + y = 7, we get x = 1, y =6]

Thus, we see that 8 is the minimum value of Z at the corner point (1, 6). Here we see that the feasible region is unbounded. Therefore, 8 may or may not be the minimum value of Z. To decide this issue, we graph the inequality

2x + y < 8.........(v)

And check whether the resulting open half has points in common with feasible region or not. If it has common point, then 8 will be the minimum value of Z, otherwise 8 will be the minimum value of Z.

Thus, from the graph it is clear that, it has no common point.

Therefore, Z = 2x + y has 8 as minimum value subject to the given constraint.

Hence, the person should take 1 unit of X tablet and 6 unit of Y tablets to satisfy the given requirements and at the minimum cost of Rs 8.

1905036_1799869_ans_fd2805bead0a4410b44cd965e4225a27.PNG

In a cattle breeding firm, it is prescribed that the food ration from one animal must contain 14.22 and 1 units of nutrients A, B, and C respectively. Two different kinds of fodder are available. Each unit of these two contains the following amounts of these three nutrients:
Fodder $\rightarrow$
Nutrient Fodder 1 Fodder 2
Nutrients A 2 1
Nutrients B
2 1
Nutrients C 1 1
The cost of fodder 1 is $\square 3$ per unit and that of fodder 2 $\square 2.$ . Formulate the LPP to minimize the cost .

Let x units of fodder 1 and y units of fodder 2 be [rescribed. The cost of fodder 1 is

$\square 3$ per unit and cost of fodder 2 is

$\square 2$ per unit.

$\therefore$ total cost is z = 3x + 3y
This is the linear function which is to be minimized. Hence it is the objective function. The constraints are as per the following table:
Fodder

$\rightarrow$

Nutrient	Fodder 1	Fodder 2	Minimum requirements
Nutrients A	2	1	14
Nutrients B	2	3	22
Nutrients C	1	1	1

From table fodder contains(2x + y) units of nutrients A, (2x + 3y) units of nutrients B and (x + y) units of nutrients C.
The minimum requirements of these nutrients are 14 units, 22 units, and 1 unit respectively.
Therefore, the constraints are

$2x + y \geq 14, 2x + 3y \geq 22, x + y \geq 1$

Since, number of units (i.e. x and y) cannot be negative, we have,

$x \geq 0, y \geq 0$
Hence, the given LPP can be formulated as
Minimize z = 3x + 2y, subject to

$2x + y \geq 14, 2x = 3y \geq22, x + y \geq 1, x \geq 0, y \geq 0$

A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B and the number of man-hours available for the firm is as follows:
Gadgets Foundry Machine shop
A 10 5
b 6 4
TIme available (Hour) 60 35
Profit on the sale of A is $\square 30$ and B is $\square 20 \;per \;units.$ Formulate the LPP to have maximum profit.

Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y. The profit on sale of A is

$\square 30$ per unit and on the sale of B is

$\square 20 \;per \;unit.$

$\therefore total \;profit \;is z = 30x + 20y$
This is a linear function that is to be maximized. Hence it is the object function. The constraints are as per the following table:

Gadgets	Foundry	Machine shop	Total available time Time (in hour)
A	10	5	60
B	6	4	35

From the time table man-hours of labour required for

$x$ units of gadgets A and y units of gadgets B in foundry is

$(10x + 6y)$ hours and total man-hours of labour required in machine shop is

$(5x +4y)$ hours.

Since the maximum time available in foundry and machine shops are 60 hours and 35 hours respectively. Therefore, the constraints are

$10x + 6y \leq 60, 5x + 4y \leq 35$
Since, x and y cannot be negative, we have

$x \geq 0, y \geq 0.$
Hence, the given LPP can be formulated as:
Maximize

$z = 30x +20,$ subject to

$10x + 6y \leq 60$

$5x + 4y \leq 35,$

$x \geq 0, y \geq 0$

Refer to ExerciseSolve the linear programming problem and determine the maximum profit to the manufacturer.

Referring to solution 12, we have minimise Z = 400x + 200y, subject to

$5x + 2y \geq 30$ .

$2x + y \geq 15, x \leq y, x \leq 0, y \leq 0$ ,

On solving x - y = 0 and 5x + 2y = 30, we get

$y = \frac{30} {7}, x = \frac{30} {7}$

On solving x - y = 0 and 2x + y = 15, we get x = 5, y = 5

So, from the shaded feasible region it is clear that coordinates of corner points are (0, 15), (5, 5) and

$\left (\frac{30} {7}, \frac{30} {7} \right)$

Hence, the minimum cost is Rs 2571.43.

1906464_1799812_ans_4a79b7366f57480bbb26499f263e0d85.png

Refer to ExerciseDetermine the maximum distance that the man can travel.

Referring to Exercise 15, we have to maximise Z = x + y,

Subject to

$2x + 3y \leq 120, 8x + 5y \leq 400, x \geq 0, y \geq 0$ .

On solving, we get 8x + 5y = 400 and 2x + 3y = 120, we get

$x = \frac {300} {7}, y = \frac {80} {7}$

From the shaded feasible region, it is clear that coordinates of corner points are (0, 0), (50, 0),

$\left( \frac {30} {7}, \frac {80} {7} \right)$ and (0, 40).

Hence, the maximum distance that the man can travel is

$54 \frac{2} {7}$ km.

1906427_1799820_ans_6ac4d85152384172afe09bcafb0c128c.png

A manufacturer produces two models of bikes- model X and model Y. Model X takes a 6 man-hours to make per unit, while model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling ans marketing costs are Rs 2000 and Rs 1000 per unit for models X and Y respectively. The total funds available for these purpose are Rs 80,000 per week. Profits per unit for models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

$\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{value of} \ Z = 1000x + 500y} \\ (0, 0) & 0 \\ (40, 0) & 40000\ (Maximum) \\ (25, 30) & 25000+ 15000 = 40000\ (Maximum) \\ (0, 45) & 22500\end{matrix}$

Let the manufacture produces x number of models X and y number of model Y bikes. Model X takes 6 man-hours to make per unit and model Y takes 10 man-hours to make per unit.

There is total of 450 man-hours available per week.

$\therefore 6x + 10y \leq 450$

$\Rightarrow 3x + 5y \leq 225$ ...........(i)

For model X and Y, handling and marketing costs are Rs 2000 and Rs 1000, respectively, total funds available for these purposes are Rs 80000 per week.

$\therefore 2000x + 1000y \leq 80000$

$\Rightarrow 2x + y \leq 80$ .......(ii)

Hence, the profits per unit for model X and Y are Rs 1000 and Rs 500, respectively,

$\therefore$ Required LPP is

Maximise Z = 1000x + 500y

Subject to,

$3x + 5y \leq 225, 2x + y \leq 80, x \geq 0, y \geq 0$

From the shaded feasible region, it is clear that coordinates of corner points are (0, 0), (40, 0), (25, 30) and (0, 45)

On solving 3x + 5y = 225 and 2x + y = 80, we get

x = 25, y = 30

So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get a maximum profit of Rs 40000.

1905024_1799857_ans_c04973038e5940a7990bfdbce43898a6.PNG

Maximise Z = x + y subject to $x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0$ .

$\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{ value of} \ Z = x + y} \\ (0, 0) & 0 \\ (3, 0) & 3 \\ \left(\dfrac{28}{11}, \dfrac{15}{11}\right) & \dfrac{43}{11} = 3 \dfrac{10}{11} \ (maximum) \\ (0,2) & 2 \end{matrix}$

Here, the given LPP is,

Maximize Z = x + y subject to,

$x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0$

On solving x + 4y = 8 and 3x + y = 9, we get

$x = \frac{28} {11}, y =\frac{15} {11}$

From the feasible region, it is clear that coordinates of corner points are (0, 0), (3, 0),

$\left ( \frac{28} {11}, \frac{15} {11} \right )$ and (0, 2)

Hence, the maximum value is

$3 \frac{10} {11}$ .

1905022_1799833_ans_9b09e614da514163acee6fb8c58fc869.PNG

Maximise and minimise Z = 3x - 4y, subject to
$x - 2y \leq 0$
$-3x + y \leq 4$
$x -y \leq 6$
$x, y \geq 0$

$\begin{matrix} \underline{\text{Corner Points}} & \underline{\text{Corresponding value of} \ Z = 3x -4y} \\ (0, 0) & 0 \\ (0, 4) & -16 \ (Minimum) \\ (12, 6) & 12 (Maximum)\end{matrix}$

Given LPP is

Maximise and minimise Z = 3x - 4y, subject to

$x - 2y \leq 0, -3x + y \leq 4, x -y \leq 6, x, y \geq 0$

[On solving x - y = 6 and x - 2y = 0, we get x = 12, y = 6]

For given unbounded region the minimum value of Z may or may not be -16. So, for deciding this, we graph the inequality.

3x - 4y < -16

And check whether the resulting open half plane has common points with feasible region or not.

Thus, from the figures it shows it has common points with feasible region, So, it does not have any minimise value.

Also, similarly for maximum value, we graph the inequality 3x - 4y > 12.

And see that resulting open has no common points with the feasible region and hence maximum value of 12 exits for Z = 3x - 4y.

1905041_1799914_ans_24fbe82a27fd4cdf883c4f602a5738b5.PNG

Solve the following LPP by graphical method:
Maximize $z = 7x + 11y,$ subject to $3x + 5y \leq 26, 5x +3y \leq 30, x \geq 0, y \geq 0$

30 respectively.

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	3x + 57 =26	$A(\dfrac{26}{3},0)$	$B(0,\dfrac{26}{5})$	$\leq$	origin side of line AB
CD	5x + 3y = 30	C(6,0)	D(0,10)	$\leq$	origin side of line CD

The feasible region is

$OCPBO$ which is shaded in the graph.
The vertices of the feasible region are O (0,0), C (6,0), P and

$B(0,\dfrac{26}{5})$ .
The vertex P is the point of intersection of the lines

$3x + 5y = 26 .........(1)$
and

$5x + 3y = 30 ..................(2)$
Multiplying equation (1) by 3 and equation (2) by 5, we get

$9x + 15y = 78$
and

$25x + 15y = 150$
On subtracting , we get

$16x = 72$

$\therefore x = \dfrac{72}{16} = \dfrac{9}{2} = 4.5$
Substitiuting

$x = 4.5$ in equation (2), we get

$5(4.5) + 3y = 30$

$22.5 + 3y = 30$

$\therefore 3y = 7.5$

$\therefore y = 2.5$

$\therefore P is (4.5, 2.5)$
The values of the objective function

$z = 11y$ at these corner points are

$z(O) = 7(0) + 11(0) = 0 + 0 = 0$

$z(C) = 7(6) + 11(0) = 42+ 0 = 42$

$z(P) = 7(4.5) + 11(2.5) = 31.5 + 27.5 = 59.0 = 59$

$z(B) = 7(0) + 11(\dfrac{26}{5}) = \dfrac{286}{5} = 57.2$

$\therefore$ z has maximum value

$59,$ when

$x = 4.5$ and

$y = 2.5.$
1988103_1849497_ans_47f2f28034ed4be997f9ce01dbb0fb6f.png

Solve then following LPP graphically:
Maximize : Z = 10x + 25y
Subjec to : $x \leq 3, y \leq 3, x + y \leq 5, x \geq 0, y \geq 0.$

FIrst we draw the lines AB, CD and EF whose equations are x = 3, y = 3 nad x + y = 5 respectively.

Line	Equation	Point on the X-axis	Point on the Y-axis	Sign	Region
AB	x =3	A(3,0)	-	$\leq$	origin side of line AB
CD	y = 3	-	D(0,3)	$\leq$	origin side of line CD
EF	x + y = 5	E(5,0)	F(0,5)	$\leq$	origin side of line EF

The feasible region is OAPQDO which is shaded in the figure.
The vertices of the feasible region are O (0,0), A 93,0), P, Q and D (0,3).
Pis the point of intersection of the lines x + y = 5, we get,

$3 + y = 5$

$y = 2$
P

$P = (3,2)$
Q is the point of intersection of the lines x + y = 5 , and y = 3
Substituting y = 3 in x+y=5, we get

$x + 3 = 5$

$x = 2$
Q = (2,3)
The values of the object function z = 10x + 25 y at these vertices are

$Z(O) = 10(0) + 25(0) = 0$

$Z(A) = 10(3) + 25(0) = 30$

$Z(P)= 10(3) + 25(2) = 30 + 50 = 80$

$Z(Q)= 10(2) + 25(3) = 20 + 75 = 95$

$Z(D) = 10(0) + 25(3) = 75$
Z has max imumvalue 95, when x = 2 and y = 3.

If john drives a car at a speed of 60 km/hour, he has to spend $\square$ 5 per km on petrol. If he drives at a faster speed of 90 km/hour, the cost of petrol increases $\square$ 8 per km. he has $\square$ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as LPP.

Let John travel

$x_{1}$ km at a speed of 60 km/hour and

$x_{2}$ km at a speed of 90 km/hour.
Therefore, time required to travel a distance of

$x_{1}$ km is

$\dfrac{x_{1}}{60} hours$ and the time required to travell a distance of

$x_{2}$ km is

$\dfrac{x_{2}}{90} hours.$

$\therefore$ total time required to travel is

$\left ( \dfrac {x_{1}}{60} + \dfrac{x_{2}}{90} \right ) hours$ .
Since he wishes to travel the maximum distance within an hour.

$\left ( \dfrac {x_{1}}{60} + \dfrac{x_{2}}{90} \right ) \leq 1$
He has to spend

$\square$ 5per km on petrol at a speed of 60 km/hour and

$\square$ 8 per km at a speed of 90 km/hour

$\therefore$ the total coist of travelling is

$\square (5x_{1} + 8x_{2})$
Since he has

$\square$ 600 to spend on petrol,

$5x_{1} + 8x_{2} = \leq 600$
Since distance is never negative

$,x_{1} \geq 0, x_{2} \geq 0.$
Total distance traveled by John is z =

$(x_{1} + x_{2})$ km.
This is the linear function which is to be maximized.
Hence, it is objective function.
Hence, the given LPP can be formulated as:
Maximize z =

$x_{1} + x_{2}$ , subject to

$\left ( \dfrac {x_{1}}{60} + \dfrac{x_{2}}{90} \right ) \leq 1, 5x_{1} + 8x_{2} = \leq 600, x_{1} \geq 0, x_{2} \geq 0$

A manufacturer produces bulbs and tubes. Each of these must be processed through two machines $M_{1}$ and $M_{2}$ . A package of build requires 1 hour of work on machine $M_{1}$ and 3 hours on machine $M_{2}$ . A package of tubes requires 2 hours on machine $M_{1}$ and 4 hours on machine $M_{2}$ . He earns a profit of $\square$ 13.5 per package of bulbs and $\square$ 55 per package of tubes. Formulate the LPP to maximize the profit, if he operates the machine $M_{1}$ , for almost 10 hours a day and machine $M_{2}$ for almost 12 hours a day.

Let the number of packages of bulbs produced by manufacturer be x and packages of tubes by y. The manufacturer earns a profit of

$\square$ 13.5 per package of bulbs and

$\square$ 55 per package of tubes.
Therefore, his total profit is p =

$\square$ (13.5x + 55y)
This is alinear function that is to be maximized, Hence, it is an objective function.
The constraints are as per the following table:

Machine	Bulbs (x)	Tubes (y)	Available time
Machine $M_{1}$	1	2	10
Machine $M_{2}$	3	4	12

From the table, the total time required for Machine

$M_{1}$ is

$(x + 2y)$ hours and for Machine

$M_{2}$ is

$(3x +4y)$ hours. Given
Machine

$M_{1}$ and Machine

$M_{2}$ are available for at most 10 hours and 12 hours a day respectively.
Therefore, the constraints are

$x + 2y \leq 10, 3x + 4y \leq 12.$ Since, x and y cannot be negative, we have,

$x \geq 0, y \geq 0.$
Hence, the given LPP can be formulated as:
Maximize

$p = 13.5x + 55y$ , subject to

$x + 2y \leq 10, 3x + 4y \leq 12, x \geq 0, y \geq 0.$

Solve the following LPP by graphical method:
Maximize $z = 4x + 6y,$ subject to $3x + 2y \leq 12, x + y \geq 4, x, y \geq 0.$

First we draw the lines AB, AD whose equations are 3x + 2y = 12 and x = y = 4 respectively.

LIne	Equation	Point on the X- axis	Point on the Y- axis	Sign	Region
AB	3x + 2y = 12	A(4,0)	B(0,6)	$\leq$	origin side of the line AB
AC	x + y = 4	A(4,0)	C(0,4)	$\geq$	non-origin side of line av

The feasible region is the

$\Delta ABC$ Which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A(4,0), B(0,6),and C(0,4).
The values of the objective function z = 4x + 6y at these vertices are

$z(A) = 4(4) + 6(0) = 16 + 0 = 16$

$z(B) = 4(0) + 6(6) = 0 + 36 = 36$

$z(C) = 4(0) + 6(4) = 0 + 24= 24$

$\therefore$ z has maximum value

$36,$ when

$x = 0$ and

$y = 6$

1993095_1849285_ans_dbc107687f4041e7b4a457fe661f84e4.png

Solve the following LPP by graphical nethod:
Maximmize : $z = 3x + 5y$
Subject to : $x + 4y \leq 24$
$3x + y \leq 21$
$x + 7 \leq 9$
$x \geq 0, y \geq 0.$

To draw the feasible region, construct table as follows:

Inequality	Corresponding Equation (of line)	Intersection of linewith X-axis	Intersection of linewith Y-axis	Region
$x + 4y \leq 24$	$x + 4y = 24$	(24,0)	(0,6)	Origin side
$3x + y \leq 21$	3x + y =21	(7,0)	(0,21)	Origin side
$x + y \leq 9$	$x + y = 9$	(9,0)	(0,9)	Origin side

Shaded portion

$OABCD$ is the feasible region,
whose vertices are

$O(0,0), A(7,0), B, C$ and

$D(0,6)$
B is the point of intersection of the lines

$3x + y = 21$ and

$x + y = 9.$
Solving the above equations, we get

$x = 6, y = 3$

$\therefore B = (6,3)$

$C$ is the point pf intersection of the lines

$x + 4y = 24$
and

$x + y = 8.$
Solving the above equations, we get

$x = 4, y = 5$

$\therefore C \equiv (4,5)$
Here, the objective function is Z = 3x + 5y,

$Z at O(0,0) = 3(0) + 5(0) = 0$

$Z at A(7,0) = 3(7) + 5(0) = 21$

$Z at B(6,3)= 3(6) + 5(3) = 18 + 15 = 33$

$Z at C(4,5)= 3(4) + 5(5) = 12 + 25 = 37$

$Z at D(0,6) = 3(0) + 5(6) = 30$

$\therefore$ Z has maximum value $37$ at $C(4,5).$

$\therefore$ Z is maximum, when $x =4, y = 5$

1993093_1849702_ans_467acc0c6ad6434384ca2cfc5fc25c27.PNG

A company manufactures two types of fertilizers $F_{1}$ and $F_{2}$ . Each type of fertilizers requires two types of raw materials A and B. The number of unit of A and B required to manufacture one unit of fertilizer $F_{1}$ and $F_{2}$ and availability of the raw materials A and B per day are given in the table below:
Fertilizers $\rightarrow$
Raw material $F_{1}$ $F_{2}$ Availability
A 2 3 40
B 1 4 70
By selling one unit of $F_{1}$ and one unit of $F_{2}$ , the company gets aprofit of $\square$ 500 and $\square$ 750 respectively. Formulate the problem as LPP to maximize the profit.

Let the company manufacturers x units of fertilizers

$F_{1}$ and y units of fertilizers

$F_{2}$ . Then the total profit to be company is z =

$\square$ (500x + 750y).
This is a linear function that is to be maximized. Hence, it is an objective function.
Fertilizers

$\rightarrow$

Raw material	$F_{1}$	$F_{2}$	Availability
A	2	3	40
B	1	4	70

The raw material A required for x units of Fertilizers

$F_{1}$ and y units of Fertilizers

$F_{2}$ is 2x + 3y. Since the maximum availability of A is 40, we have the first constraints as

$2x + 3y \leq 40$
Similarly, considering the raw material B, we have

$x + 4y \leq 70$
Since, x and y cannot be negative, we have,

$x \geq 0, y \geq 0.$
Hence, the given LPP can be formulated as:
Maximize z = 500x +750 y, subject to

$2x + 3y \leq 40, z + 4y \leq 70, x \geq 0, y \geq 0.$

The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be at least 5 kg. Cement costs $\square$ 20 per kg and sand costs of $\square$ 6 per kg. Strength consideration dictates that a concrete brick should contain minimum 4 kg of cement and not more than 2 kg of sand. Form the LPP for the cost to be minimum.

Let the company use

$x_{1}$ kg of cement and

$x_{2}$ kg of sand to make concrete bricks.
Cement cost

$squqre$ 20 per kg and sand cost

$\square$ 6 per kg.

$\therefore$ the total cost c =

$\square (20x_{1} + 6x_{2})$
This is a linear function which is to be minimized
Hence, it is an objective function.
Total weight of brick =

$(x_{1} + x_{2}) kg$
Since, the weight of concrete bricks has to be at least 5 kg,

$\therefore x_{1} + x_{2} \leq 5$
Since concrete brick should contain minimum 4 kg of cement and not more than 2kg of sand,

$x_{1} \geq 4 \;and \;0 \leq x_{2} \leq 2$
Hence, the given LPP can be formulated as:
Minimize

$c = 20x_{1} + 6x_{2}$ , subject to

$x_{1} + x_{2} \geq 5, x_{1} \geq 4, 0 \leq x_{2} \leq 2$

A doctor has prescribed two different units of food A and B to form a weekly diet for a sick person. The minimum requirements of fat, carbohydrates and proteins are 18, 24, 14 units respectively. One unit of food A has 4 units of fat, 14 units of carbohydrates and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is $\square$ 4.5 per unit and that of food B is $\square$ 3.5 per unit. Form the LPP, so that the sick person's diet meets the requirements at a minimum cost.

Let the diet of sick person include x units of food A and y units of food B.
Then

$x \geq 0, y \geq 0.$
The prizes of food A and B are

$\square$ 4.5 and

$\square$ 3.5 per unit respectively.
Therefore, the total cost is z =

$\square$ (4.5x + 3.5y)
This is the linear function that is to be minimized. Hence, it is an objective function. The constraints are as per the following table:
Food type

$\rightarrow$

Ingredients	Food A (x)	Food B (y)	Minimum requirements
Fats	4	6	18
Carbohydrates	14	12	28
Proteins	8	8	14

From the table, the sick person's diet will include

$(4x + 6y)$ units of fats,

$(14x + 12y)$ units of carbohydrates and

$(8x + 8y)$ units of proteins. The minimum requirements of these ingredients are

$18$ units,

$28$ units, and

$14$ units respectively.
Therefore, the constraints are

$4x + 6y \geq 18, 14x + 12y \geq 28, 8x + 8x \geq14, x \geq 0, y \geq 0.$
Hence the given

$LPP$ can be formulated as
Minimize

$z = 4.5x + 3.5y,$ subject to

$4x + 6y \geq 18, 14x + 12y \geq 28, 8x + 8x \geq14, x \geq 0, y \geq 0.$

A printing company prints two types of magazines A and B. The company earns $\square$ 10 and $\square$ 15 in magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on Machine I, 5 hours on Machine II, and 2 hours on Machine III. Magazine B requires 3 hours on Machine I, 2 hours on Machine II, and 6 hours on Machine III. Machines I, II, III are available for 36,50, and 60 hours per week respectively. Formulate the LPP to determine weekly production of magazines A and B, so that the total profit is maximum.

Let the company print x magazine of type A and y magazine of type B.

Profit on scale of magazine A is

$\square$ 10 per copy and magazine B is

$\square$ 15 per copy. Therefore, the total earning z of the company is z =

$\square$

$(10x + 15y).$

This is linear function that is to be maximized. Hence, it is an objective function,

The constraints are as per the following table:

Magazine type

$\rightarrow$

Magazine type Time required per unit

Magazine A (x) Time required per unit

Magazine B (y) Available time per week (in hours)

Machine I 2 3 36

Machine II 5 2 50

Machine III 2 6 60

From the table, the total time required for Machine I is

$(2x + 3y)$ hours, for Machine II is

$(5x + 2y)$ hours and Machine III is

$(2x + 6y)$ hours.

The Machines I,II,III are available for 36, 50, and 60 hours per week. Therefore, the constraints are

$2x + 3y \leq 36, 5x + 2y \leq 50, 2x + 6y \leq 60.$

Since x and y cannot be negative. We have,

$x \geq 0, y \geq 0.$ Hence, the given LPP can be formulated as:

Maximize

$z = 10x + 15y,$ subject to

$2x + 3y \leq 36, 5x + 2y \leq 50, 2x + 6y \leq 60, x \geq 0, y \geq 0.$

Solve the following LPP:
Maximize $z = 4x_{1} + 3x_{2}$ subject to
$3x_{1} + x_{2} \leq 15, 3x_{1} + 4x_{2} \leq 24, x_{1} \geq 0, x_{2} \geq 0.$

First
we draw the line AB and CD whose equations are $3x_{1} + x_{2} = 15$
and $3x_{1} + 4x_{2} = 24$ respectively

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	$3x_{1} + x_{2} = 15$	A(5,0)	B(0,15)	$\leq$	origin side of line AB
CD	$3x_{1} + 4x_{2} = 24$	C(8,0)	O(0,6)	$\leq$	origin side of the line CD

The Feasible region is OAPDO which is shaded in the graph.

The
vertices of the feasible region are O(0,0), A(5,0),P and D (0,6).

P is the point of intersection of the lines

$3x_{1} + 4x_{2} = 24$ .............(1)

and $3x_{1} + x_{2} = 15$ ...........(2)

On subtracting, we get,

$3x_{2} = 9$

$\therefore x_{2} = 3$

Substituting $x_{2} = 3$ in (2), we get

$3x_{1} + 3 = 15$

$\therefore 3x_{1} = 12$

$\therefore x_{1} = 4$

$\therefore P \;is (4,3 )$

The value of objective function $z = 4x_{1} + 3x_{2}$ at these vertices are

$z(O) = 4(0) + 3(0) = 0 + 0 = 0$

$z(A) = 4(5) + 3(0) = 20 + 0 = 20$

$z(P) = 4(4) + 3(3) = 16 + 9 = 25$

$z(D) = 4(0) + 3(6) = 0 + 18 = 18$
$\therefore$ z has maximum value $25$ when $x = 4$ and $y = 3.$

1992972_1850382_ans_d1b8813181ae4fccb3e8bdf1faf9e1d7.png

Solve then following LPP graphical method:

Maximize : Z = 7x + y

Subject to : $5x +y \geq 5, x + y \geq 3, x \geq 0, y \geq 0.$

FIrst we draw the lines AB, CD and EF whose equations are x = 3, y = 3 nad x + y = 5 respectively.

Line	Equation	Point on the X-axis	Point on the Y-axis	Sign	Region
AB	x =3	A(3,0)	-	$\leq$	origin side of line AB
CD	y = 3	-	D(0,3)	$\leq$	origin side of line CD
EF	x + y = 5	E(5,0)	F(0,5)	$\leq$	origin side of line EF

The feasible region is OAPQDO which is shaded in the figure.
The vertices of the feasible region are O (0,0), A 93,0), P, Q and D (0,3).
Pis the point of intersection of the lines x + y = 5, we get,

$3 + y = 5$

$y = 2$
P

$P = (3,2)$
Q is the point of intersection of the lines x + y = 5 , and y = 3
Substituting y = 3 in x+y=5, we get

$x + 3 = 5$

$x = 2$
Q = (2,3)
The values of the object function z = 10x + 25 y at these vertices are

$Z(O) = 10(0) + 25(0) = 0$

$Z(A) = 10(3) + 25(0) = 30$

$Z(P)= 10(3) + 25(2) = 30 + 50 = 80$

$Z(Q)= 10(2) + 25(3) = 20 + 75 = 95$

$Z(D) = 10(0) + 25(3) = 75$
Z has max imumvalue 95, when x = 2 and y = 3.

Solve the following LPP:
Maximize $z = 5x_{1} + 6x_{2}$ subject to $2x_{1} + 3x_{2} \leq 18, 2x_{1} + x_{2} \leq 12, x_{1} \geq 0, x_{2} \geq 0.$

First we draw the line AB and CD whose equations are

$2x_{1} + 3x_{2} = 18$ and

$2x_{1} + x_{2} = 12$ respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	$2x_{1} + 3x_{2} = 18$	A(9,0)	B(0,6)	$\leq$	origin side of line AB
CD	$2x_{1} + x_{2} = 12$	C(6,0)	O(0,12)	$\leq$	origin side of the line CD

The Feasible region isOCPBO which is shaded in the graph. The vertices of the feasible region areO(0,0), C(6,0),P and B (0,6).
P is the point of intersection of the lines

$2x_{1} + 3x_{2} = 18$ .............(1)
and

$2x_{1} + x_{2} = 12$ ...........(2)
On subtracting, we get,

$2x_{2} = 6$

$\therefore x_{2} = 3$
Substituting

$x_{2} = 3$ in (2), we get

$2x_{1} + 3 = 12$

$\therefore x_{1} = 9$

$\therefore P \;is \left ( \dfrac{9}{2},3 \right )$
The value of objective function

$z = 5x_{1} + 6x_{2}$ at these vertices are

$z(O) = 5(0) + 6(0) = 0 + 0 = 0$

$z(C) = 5(6) + 6(0) = 30 + 0 = 30$

$z(P) = 5 \left ( \dfrac{9}{2} \right ) + 6(3) = \dfrac{45}{2} + 18 = \dfrac{45 + 36}{2} = \dfrac{81}{2} = 40.5$

$z(B) = 5(0) + 6(3) = 0 + 18 = 18$
Maximum value of

$z$ is

$40.5$ when

$x_{1} = \dfrac{9}{2}y_{1} = 3.$

1992979_1850255_ans_231145f10779424c9b11a273a2fce66a.png

Solve the following LPP by graphical method:
Mininize z = 8x + 10y,
Subject to $2x + y \geq 7, 2x + 3y \geq 15, y \geq 2, x \geq 0, y \geq 0.$

FIrst
we draw the lines $AB, CD$ and $EF$ whose equations are $2x + y = 7, 2x + 3y = 15$ and $x + y = 5$
respectively.

Line	Equation	Point on the X-axis	Point on the Y-axis	Sign	Region
AB	2x + y = 7	A(3,5,0)	B(0,7)	$\geq$	non-origin side of line AB
CD	2x =3y = 15	C(7,5,0)	D(0,5)	$\geq$	non-origin side of line CD
EF	y = 2	-	F(0,2)	$\geq$	non-origin side of line EF

The feasible region is $EPQBY$ which is shaded in the graph.

The vertices of the feasible region are $P, Q$ and $B (0,7), P$ .
Substituting $y = 2$ in $2x+3y=15,$ we get

$2x + 3y = 15$

$\therefore 2x = 9$
$\therefore x = (4.5)$

$\therefore P = (4.5, 2)$

Q is the point of intersection of the lines

$2x +3y =15...............(1)$

and $2x + y = 7$

On subtracting, we get

$2y = 8$

$\therefore y = 4$

$\therefore$ from (2), 2x + 4 =7

$\therefore 2x =3$

$\therefore x = 1.5$

$\therefore Q = (1.5, 4)$

Z has minimum value $52,$ when $x = 5$ and $y = 4$ .

1992987_1849792_ans_9a08b181024b4bd6a3f0832900ea87cf.png

Solve then following LPP graphical method:

Maximize : z = 6x + 21y

Subject to : $x + 2y \geq 3, x + 4y \geq 4, 3x + y \geq 3, x \geq 0, y \geq 0.$

FIrst
we draw the lines AB, CD and EF whose equations are x +2y = 3, x + 4y = 4 and 3x + y = 3 respectively.

Line	Equation	Point on the X-axis	Point on the Y-axis	Sign	Region
AB	x +2y =3	A(3,0)	$B(0,\dfrac{3}{2})$	$\geq$	non-origin side of line AB
CD	x +4y = 4	C (4,0)	$D(0,1)$	$\geq$	non-origin side of line CD
EF	3x + y = 3	E(1,0)	$F(0,3)$	$\geq$	non-origin side of line EF

The
feasible region is

$XCPQFY$ which is shaded in the graph.

The vertices of the feasible region are

$C (4,0), P, Q$ and

$F (0,3).$

P is the point of intersection of the lines

$x + 4y= 4$ and

$x + 27 = 3,$ we get,
On subtracting, we get

$2y =1$

$\therefore y = \dfrac{1}{2}$
Substituting y =

$\dfrac{1}{2}$ in x + 2y = 3, we get

$x + 2(\dfrac{1}{2}) = 3$

$\therefore x = 2$

$\therefore P \equiv (2, \dfrac{1}{2})$
Q is the point of intersection of these lines

$x + 2y = 3$ ..............(1)
and

$3x + y = 3$ ................(2)
Multiplying equation (1) by 3 , we get

$3x + 6y = 9$
Subtracting equation (2) from the equation, we get

$5y = 6$

$\therefore y = \dfrac{6}{5}$

$\therefore$ from...(1),

$x + 2(\dfrac{6}{5}) = 3$

$\therefore x = 3 - \dfrac{12}{5} = \dfrac{6}{5}$

$\therefore Q \equiv (\dfrac{3}{5},\dfrac{6}{5})$
The values of the objective function

$z = 6x + 21y$ at these vertices are

$z(C) = 6(4) + 21(0) = 24$

$z(P) = 6(2) + 21(\dfrac{1}{2})$

$= 12 + 10.5 = 22.5$

$z(Q) = 6(\dfrac{3}{5}) + 21(\dfrac{6}{5})$

$= \dfrac{18}{5} + \dfrac{126}{5} = \dfrac{144}{5} = 28.8$

$z(F) = 6(0) + 21(3) = 63$

$\therefore z$ has minimum value

$22.5,$ when

$x = 2$ and

$y = \dfrac{1}{2}.$

1988096_1849852_ans_00682a0b95424eaaabf270864b0091bf.png

Solve the following problem graphically:
Minimize $Z = -3x + 4y$ subject to $x + 2y \leq 8; 3x + 2y \leq 12; x \geq 0, y \geq 0.$

$x + 2y = 8$

$X$	$2$	$4$	$0$
$Y$	$3$	$2$	$4$

$0 + 0 < 8$
true towards

$0$

$3x + 2y = 12$

$X$	$2$	$4$	$0$
$Y$	$9$	$0$	$6$

$0 + 0 < 12$
true shaded towards

$0$
ABCD is the solution region

$x + 2y = 8$
Put

$x = 0 \Rightarrow A(-0, 4)$
Put

$y = 0 \Rightarrow B(8, 0)$
At

$A(0, 4) \Rightarrow Z = 16$
At

$B(2, 3) \Rightarrow Z = -3 \times 2 + 4 \times 3 = 6$
At

$C(4, 0) \Rightarrow Z = -12$
At

$D(0, 0) \Rightarrow Z = 0$
Hence Z is minimum at

$C(4, 0) = -12$
1955542_1859926_ans_561bc244243844bc92e41bbe2bad8d1b.png

Solve the following problem graphically:
Maximize $Z = 3x + 4y$ subject to the constraints : $x + y \leq 4, x \geq 0, y \geq 0.$

$x + y = 4$

$X$	$2$	$3$	$1$
$Y$	$2$	$1$	$3$

put

$x = 0, y = 0$
since

$0 + 0 \leq 0,$ equation is true.

$\therefore$ This will be shaded towards

$0.$
At

$A(4, 0), Z = 3 \times 4 = 12$
At

$B(0, 4) , Z = 4 \times 4 = 16$

$\therefore Z$ is maximised at

$B(0, 4) = 16$
1866539_1859923_ans_7cbc32809d074f06ab816495fe528611.png

Show that the minimum of Z occurs at more than two points
Minimise and Maximise $Z = x + 2y$
subject to $x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200; x, y \geq 0.$

$x + 2y = 100$

$x$	$100$	$50$	$20$
$y$	$0$	$20$	$40$

$0 + 0 \ngeq 100$

$0 - 0 < 0$ False

$2x - y = 0$

$x$	$100$	$20$	$10$
$y$	$200$	$40$	$20$

$0 - 0 < 0$
False

$2x + y = 200$

$x$	$100$	$50$	$60$
$y$	$0$	$100$	$80$

$0 + 0 < 200$
True
ABCD is the solution
At

$A(0, 200), Z = 2 \times 200 = 400$
At

$B(50, 100), Z = 50 + 2 \times 100 = 250$
At

$C(20, 40), Z = 20 + 2 \times 40 = 100$
At

$D(0, 50), Z = 2 \times 50 = 100$
Z is maximised at

$A(0, 200) = 400$
Z is minimised at

$2$ points

$C(20, 40) \& D(0, 50) = 100$
1955548_1859938_ans_c31b5a25770c472199e562607dd48211.png

Show that the minimum of Z occurs at more than two points
Maximise $Z = x + y,$
subject to $x - y \leq -1, -x + y \leq 0, x, y \geq 0.$

$x - y = -1$

$x$	$1$	$0$	$2$
$y$	$2$	$1$	$3$

$0 - 0 \ngeq -1$
False

$x + 2y = 6$

$x$	$0$	$4$	$2$
$y$	$3$	$1$	$2$

$0 + 0 \nless 0$
False,
Since there is no common shaded region, there is no feasible solution region. Hence maximum of Z does not exist.

A factory makes tennis rackets and cricket bats. A tennis racket takes $1.5$ hours of machine time and $3$ hours of craftman's time in its making while a cricket bat takes $3$ hour of machine time and $1$ hour of craftman's time. In a day, the factory has the availability of not more than $42$ hours of machine time and $24$ hours of craftman's time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity ?
(ii) If the profit on a racket and on a bat is $Rs.20$ and $Rs.10$ respectively, find the maximum profit of the factory when it works at full capacity.

Machine time	Craftman's time
Racket (x)	$1.5 \,hrs = \dfrac{3}{2}$	$3 \,hrs$
Bat(y)	$3 \,hrs$	$1 \,hr$
Total	$42 \,hrs$	$24 \,hrs$

$Z = x + y \,P = 20x + 10y$
Z subject to constraints
(i)

$\dfrac{3}{2}x + 3y \leq 42$

$x + 2y \leq 28$
(ii)

$3x + y \leq 24$
(iii)

$x \geq 0; y \geq 0$

$x + 2y = 28$

$x$	$4$	$2$
$y$	$12$	$13$

$3x + y = 24$

$x$	$4$	$6$
$y$	$12$	$6$

ABCD is the solution region

$A(0, 0), Z = 0, P = 0$

$B(8, 0), Z = 8, P = 20 \times 8 = Rs.160$

$C(4, 12), Z = 16, P = Rs.200$

$D(0, 14), Z = 14, P = Rs.140$
At

$C(4, 12)$ Z is maximised to

$16$ and then profit is also maximised by the manufacture of

$4$ rackets and

$12$ bats.

1955557_1859957_ans_25f07f5f34184ce4935689fcf099d2c3.png

A manufacturer produces nuts and bolts. It takes $1$ hour of work on machine A and $3$ hours on machine B to, produce a package of nuts. It takes $3$ hours on machine A and $1$ hour on machine B to produce a package of bolts. He earns a profit of ? $17.50$ per package on nuts and ? $7.00$ per package on bolts. How many packages of each should be produced each day his machines for at the most $12$ hours a day ?

	Nut (x)	Bolt (y)
Machine A	$1 \,hr$	$3 \,hrs$
Machine B	$3 \,hrs$	$1 \,hr$
Profit	$17.50$	$7.00$

$P = \dfrac{35}{2}x + 7y$
(i)

$x + 3y \leq 12$
(ii)

$3x + y \leq 12$
(iii)

$x \geq 0, y \geq 0$

$x + 3y = 12$

$x=$	$0$	$,3$	$,6$
$y=$	$4$	$,3$	$,2$

$3x + y = 12$

$x=$	$4$	$,3$	$,2$
$y=$	$0$	$,3$	$,6$

ABCD is the solution region

$A(0, 0) P = 0$

$A(0, 0) \,p = 0$

$B(4, 0) \,P = Rs.70.$

$C(3, 3) \,P = Rs.73.5$

$D(0, 4) \,P = Rs.28$
At

$(3, 3)$ he can maximise his profit. By producing

$3$ packages of nuts and

$3$ packages of bolts each.
1995827_1859960_ans_4a5cad92c60d478eb5cc01b7963dad01.png

Solve the following LPP:
Maximize z = 60x + 50 y subject to
$x + 2y \leq 40, 3x + 2y \leq 60, x \geq 0, y \geq 0.$

First
we draw the lines AB and CD whose equations are $x + 2y = 40$
and $3x + 2y = 60$ respectively

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	$x + 2y = 40$	A(40,0)	B(0,20)	$\leq$	origin side of line AB
CD	$3x + 2y = 60$	C(20,0)	O(0,30)	$\leq$	origin side of the line CD

The Feasible region is OCPBO which is shaded in the graph.

The
vertices of the feasible region are O(0,0), C(20,0),P and B (0,20).

P is the point of intersection of the lines

$3x + 2y = 60$ .............(1)

and $x + 2y = 40$ ...........(2)

On subtracting, we get,

$2x = 20$

$\therefore x = 10$

Substituting $x = 10$ in (2), we get

$10 + 2y = 40$

$\therefore 2y = 30$

$\therefore y = 15$

$\therefore$ P is (10, 15)

The value of objective function $z = 60x + 50y$ at these vertices are

$z(O) = 60(0) + 50(0) = 0 + 0 = 0$

$z(C) = 60(20) + 50(0) = 1200 + 0 = 1200$

$z(P) = 60(10) + 50(15) = 600 + 750 = 1350$

$z(B) = 60(0) + 50(20) = 0 + 1000 = 1000$

$\therefore$ z has maximum value $1350$ at $x = 10, y = 15$

1992968_1850413_ans_af1e08e25aa440b58c1b5f09abbcac13.png

Show that the minimum of Z occurs at more than two points
Maximise $Z = -x + 2y,$ subject to the constraints:
$x \geq 3, x + y \geq 5, x + 2y \geq 6, y \geq 0$

$x + y = 5$

$x$	$1$	$3$	$2$
$y$	$4$	$2$	$3$

$0 + 0 \ngeq 5$
False

$0 \ngeq 3$
False

$x + 2y = 6$

$x$	$0$	$4$	$2$
$y$	$3$	$1$	$2$

$0 + 0 \ngeq$
False
ABC is the solution region
At

$A(3, 2) , Z = -3 + 2 \times 2 = 1$
At

$B(4, 1), Z = -4 + 2 \times 1 = -2$
At

$C(6, 0), Z = -6 = -6$
Z is maximised at

$A(3, 2) = 1$ and
minimised at

$C(6, 0) = -6$
1955554_1859940_ans_c0f804c3421f49f2abb5f96df1f89d88.png

Solve the following problem graphically:
Minimum $Z = 3x + 5y$ such that $x + 3y \leq 3, x + y \geq 2, x, y \geq 0.$

$x + 3y = 3$

$X$	$3$	$0$ $
$Y$	$0$	$1$

$o + 0 \ngeq 3$
false

$\therefore$ shading away from

$0$

$x + y = 2$

$X$	$0$	$2$	$1$
$Y$	$2$	$0$	$1$

$0 + 0 \ngeq 2$
false
shading away from

$0$
Away from

$0$
corners points of feasible region are

$A(0, 2); Z = 5 \times 2 = 10$

$B(1.5, 0.5); Z = 7$

$C(3, 0); Z = 3 \times 3 = 9$

$\therefore$ Z is minimum at

$B(1.5, 0.5) = 7.$
1955544_1859931_ans_2b6519d0792b486f967da2d2bc7dd9d0.png

One kind of cake requires $200\,g$ of flour and $25\,g$ of fat, and another kind of cake requires $100\,g$ of flour and $50\,g$ of fat. Find the maximum number of cakes which can be made from $5 \,kg$ of flour and $1 \,kg$ of fat assuming that there is no shortage of the other ingredients used in making the cakes.

	I(x)	II(y)
Flour	$200\,g$	$100\,g$
Fat	$25\,g$	$50 \,g$

$Z = x + y$ subject to the constraints
(i)

$200x + 100y \leq 5000 \Rightarrow 2x + y < 50$
(ii)

$25x + 50y \leq 1000 \Rightarrow x + 2y < 40$
(iii)

$x, y \leq 0$

$2x + y = 50$

$x$	$10$	$20$	$25$
$y$	$30$	$10$	$0$

$x + 2y = 40$

$x$	$0$	$20$
$y$	$20$	$10$

ABCD is the solution region

$A(0, 0), Z = 0$

$B(25, 0)$

$Z = 25 + 0 = 25$

$C(20, 10)$

$Z = 20 + 10 = 30$

$D(0, 20)$

$Z = 0 + 20 = 20$
At

$C(20, 10)$ maximum number of cakes can be made.
1955555_1859947_ans_139c845454af478d910ad34a4e76a323.png

Minimize $Z = - 50x + 20y$
Subject to the constraints
$2x - y \geq -5$
$3x + y \geq 3$
$2x- 3y \leq 12$
and $x \geq 0, y \geq 0$

Converting the given inequations into equations

$2x – y = - 5$ …..(1)

$3x + y = 3$ …..(2)

$2x - 3y = 12$ …..(3)

Region represented by

$2x – y \geq – 5$ : The line

$2x – y = -5$ meets the coordinate axis at

$A(,0)$ and

$B(0,5)$

$2x – y = -5$

X -5/2 0

y 0 5

$A(,0) ;B (0,5)$

Join the points A and B to obtain the line. Clearly

$(0, 0)$ satisfies the inequation

$2x – y \geq – 5$ . So the region containing the origin, represents the solution set of the inequation.

Region represented by

$3x + y \geq 3$ : The line

$3x + y = 3$ meets the coordinate axis at

$A(1, 0)$ and

$B(0, 3)$ .

$3x + y = 3$

X 1 0

y 0 3

$C(1,0);D(0,3)$

Join the points C to D to obtain the line. Clearly the point

$(0, 0)$ does not satisfy the inequation. So the region opposite to the origin represents the solution set of the inequation.

Region represented by

$2x – 3y \leq 12$ : The line

$2x - 3y =12$ meets the coordinate axis at

$E(6, 0)$ and

$F(0, -4)$ .

$2x – 3y = 12$

X 6 0

y 0 -4

$E(6,0);F(0,-4)$

Join the points E and F to obtain the line. Clearly (0, 0) satisfies the inequation

$2x – 3y \leq 12$ . So the region containing the origin represents the solution set of the inequation.

Region represented by

$x \geq 0, y \geq 0$ : Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations

$x \geq 0, y \geq 0$ .

The point of intersection of lines

$2x - y = -5$ and

$3x + y = 3$ is G (,).

The shaded area is an open and common region which satisfies the given constraints. This is the proper solution of the given LPP.

The value of the objective function on These points are given in following table:

Point x Coordenate y Coordinal

$z = -50x + 2y$

B 0 5

$z_B = -50 \times 0 + 20 \times 5 = 100$

D 0 3

$z_D = -50 \times 0+ 20 \times 3 = 60$

C 1 0

$z_C = -50 \times 1 + 20 \times 0 = -50$

E 6 0

$z_E = -50 \times 6 + 20 \times 0 = -300$

Clearly Z is minimum at

$x = 6$ and

$y = 0$

∵ Minimum value of

$Z = - 300$

1866547_1872717_ans_db0fc9e5695244148cbfef6bfd8b3dc9.PNG

Minimize $Z = -3x + Ay$
$3x + 2y \leq 12$
and $x \geq 0, y \geq 0$

Converting the given inequations into equations
$x + 2y = 8$
$3x + 2y = 12$
Region represented by $x + 2y \leq 8$ : The line $x + 2y = 8$ meets the coordinate axis at $A(8, 0)$ and $B(0, 4)$ .
$x + 2y = 8$

X	8	0
y	0	4

$A(8,0),B(0,10)$
Join the points A and B to obtain the line. Clearly

$(0, 0)$ satisfies the inequation

$x + 2y = 8$ . So the region containing the origin represents the solution set of the inequation.

Region represented by $3x + 2y \leq 12$ : The line $3x + 2y =12$ meets the coordinate axis at $C(4, 0)$ and $D(0, 6)$ . $3x + 2y = 12$

X	4	0
y	0	6

$C(4, 0); D(0,6)$
Join the points C and D to obtain the line. Clearly $(0, 0)$ satisfies the inequation $3x + 2y = 12$ . So the region containing. The origin represents the solution set of the inequations.

Region represented by $x \geq 0, y \geq 0$ : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations $x \geq 0, y \geq 0$ .

The shaded region OCEB represents the common region of the above inequations. This region is the feasible region of the given linear programming problem.

The coordinates of the comer points of the feasible region are $O(0, 0), C(4, 0), E(2, 3)$ and $5(0, 4).$
The point $E(2, 3)$ has been obtained by solving the equations of the corresponding intersecting lines simultaneously.
The values of the objective function on these points are given in the following table :

Point	x-coordinate	y-coordinate	Objective function $Z = 3x + 4y$
O	0	0	$Z_0 = -3(0) + 4(0) = 0$
C	4	0	$Z_c = -3(4) + 4(0) = -12$
E	2	3	$Z_E = -3(2) + 4(3) = 6$
B	0	4	$Z_B = -3(0) + 4(4) = 16$

It is clear from the above table that the objective function has minimum value at comer point $C(4, 0)$ . So the minimum value of given linear programming problem at $x = 4$ and $y = 0$ is $- 12$ .

1866541_1872711_ans_68c6f40892d9441389a989ee431d1b2a.PNG

Find graphically, the maximum value of $z = 2x + 5y$ , subject to constraints given below :
$2x + 4y \leq 8$
$3x + y \leq 6$
$x + y \leq 4$
$x \geq 0 , y \geq 0$

Maximise

$z = 2x + 5y$ , Subject to the costraints

$2x + 4y \le 8 \Rightarrow x + 2y \le 4$

$3x + y \le 6 , x + y \le 4, x \ge 0 , y \ge 0$ .

Draw the lines

$x + 2y = 4$ (passes through

$4, 0), (0,2));$

$3x + y = 6$ (passes through

$(2, 0), (0,6)$ ) and

$x + y = 4$ (passes through

$(4, 0), (0,4)$ ).

Shade the region satisfied by the given inequations;

The shaded region in the figure gives the feasible region determined by the given inequations.

Solving

$3x + y = 6$ and

$x + 2y = 4$ simultaneously, we get

$x = \dfrac{8}{5}$ and

$y = \dfrac{6}{5}$

We observe that the feasible region

$OABC$ is a convex polygon and bounded and has corner points.

$O(0,0) , A (2, 0), B \left(\dfrac{8}{5} , \dfrac{6}{5} \right) , C(0,2)$

The optimal solution occurs at one of the corner points.

$O(0,0) , z = 2.0 + 5.0 = 0$ ;

$A (2, 0) , z = 2.2 + 5.0 = 4$ ;

$B \left(\dfrac{8}{5} , \dfrac{6}{5}\right) , z = 2. \dfrac{8}{5} + 5. \dfrac{6}{5} = \dfrac{46}{5}$ ;

$C(0,2), z = 2.0 + 5.2 = 10$ ;

Therefore, z maximum value at C and maximum value

$= 10$

1965107_1944566_ans_01fb19759e9840a49a6ecade93f5a81d.png

One kind of cake required $300 \,g$ of flour and $15\,g$ of fat, another kind requires $150 \,gm$ of flour and $30 \,gm$ of fat. Find the maximum number of cakes which can be made from $7.5 \,kg$ of flour and $600\,gm$ of fat.

1959479_1860064_ans_85ff3314831f49908b171eb00e65842f.png

The marks obtained by a student of class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.

Let

$x$

$be\space the\space marks\space obtained\space by\space student\space in\space the\space annual\space examination\space$ .
Then

$\dfrac{62 + 48 + x}{3} \geq 60$

$\Rightarrow 110 + x \geq 180$

$\Rightarrow x \geq 70$

$Thus\space, the\space student\space must\space obtain\space a\space minimum\space of\space 70\space marks\space to\space get\space an\space average\space of\space at\space least\space 60\space marks$ .

Minimize $Z = x + 2y$
Subject to the constraints
$2x + y \geq 3$
$x + 2y \geq 6$
and $x \geq 0,\,y \geq 0$

Converting the given inequations into equations
$2x + y = 3$ ….(1)
$x + 2y = 6$ ….(2)
Region represented by $2x + y \geq 3$ : The line $2x + y - 3$ meets the coordinate axis at points $A(,0)$ and $B(0. 3)$ .
$2x + y = 3$

X	3/2	0
y	0	3

$A(,0);B(0,3)$

Join the points, A and B to get the line. Clearly $(0,0)$ does not satisfy the inequation $2x + y \geq 3$ . So the region opposite to the origin represents the solution set of the inequation.

Region represented by $x + 2y \geq 6$ : The line $x + 2y = 6$ meets the coordinate axis at point $C(6, 0)$ and $D(0,3)$ .
$x + 2y = 6$

X	6	0
y	0	3

$C(6, 0);\,D(0, 3)$

Join point C to D to obtain the line. Clearly $(0,0)$ does not satisfy the inequation $x + 2y \geq 6$ . So the region opposite to the origin represents the solution set of the inequation.

Region represented by $x \geq 0$ and $y \geq 0$ : Since every point in the first quadrant satisfies the inequations. So the first quadrant is the region represented by the inequations $x \geq 0$ and $y \geq 0$ .

In the shaded region on CB, each and every point on line CB is satisfying the given inequations.
So the values of the objective function on these points are given in the following table :

Point	x-coordinate	v-coordinate	Objective function $Z = x + 2y$
5	0	3	$Z_B= 0 + 2(3) = 6$
C	6	0	$Z_c = 6 + 2(0) = 6$

It is clear from the table that required solution of given L.P.P. every point is on line BC and the minimum value of $Z = 6$ .

1866551_1873221_ans_5fc9bc3f03ba4eeaa0de11594cf25d3e.PNG

Maximize $Z = x + y$
Subject to the constraints
$x - y \leq -1$
$-x + y \leq 0$
and $x \leq 0,y \leq 0$

Converting the given inequations into equations
$x - y = - 1$
$-x + y = 0$
Region represented by $x - y \leq -1$ : Following are the points obtained by the line
$x - y = -1$

X	-1	0
y	0	1

$A(-1,0);B(0,1)$
Join the points A and B to obtain the line. Clearly $(0, 0)$ does not satisfy the inequation $x - y \leq -1$ . So the region opposite to the origin represent the solution set of the inequation.

Region represented by $-x + y \leq 0$ : The line $-x + y = 0$ meets the coordinates axis at $B(0,0)$ and $C( 1,1)$ .
$-x + y = 0$

X	0	1
y	0	1

$O(0,0);C(1,1)$
Join the points O and C to obtain this line.
Clearly (0,0) satisfies the inequation $-x + y \leq 0$ . So, the region containing the origin represents the solution set of this inequation.

Region represented by $x \geq 0, y \geq 0$ : Since every point in the quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations $x \geq 0$ and $y \geq 0$ .

It is clear from the figure that line joining point $A(-1, 0)$ and $B(0, 1)$ is parallel to the line joining point $C(1,1)$ to $D(2, 2)$ . Hence there is no finite solution of the given LPP. Minimum and Maximum value do not exist.

1866554_1873232_ans_01b0d28e3a4d41849f9ae2f219fb4ce5.PNG

Maximize
$Z = -x + 2y$
Subject to the constraints
$x \geq 3$
$x + y \geq 5$
$x + 2y \geq 6$ and
$x \geq 0,y \geq 0$

Converting the given inequations into equations
$x = 3$ …..(1)
$x +y = 5$ …..(2)
$x + 2y = 6$ …..(3)
$y = 0$ …..(4)
Region represented by $x + y \geq 5$ : The line $x + y = 5$
meets the coordinate axis on points $A(5,0)$ and $B(0, 5)$ .
$x + y = 5$

X	5	0
y	0	5

$A(5,0);B(0,5)$
Join the points A to $5$ to obtain the line. Clearly $(0, 0)$ does not satisfy the inequation $0 + 0 = 0 \geq 5$ . So the opposite region to the origin represents the feasible solution region.

Region represented by $x + 2y \geq 6$ : The line $x + 2y = 6$ meets the coordinate axis at point $C(6,0$ ) and $D(0, 3)$ .
$x + 2y = 6$

X	6	0
y	0	3

$C(6,0);D(0,3)$
Join point C to D to obtain the line. Clearly $(0,0)$ does not satisfy the inequation $0 + 2(0) = 0 \geq 6$ . So the region opposite to the origin represents the solution region of the inequation.

The region represented by $x \geq 3$ and $y \geq 0$ : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations $x\geq 3$ and $y \geq 0$ .

It is clear from the graph that there is no common region of the given inequations.
Hence there does not exist any maximum value for the given inequations.

1866559_1873249_ans_0f73e9cc1bc14bbab70acf445c69d758.PNG

Two tailors, A and B, earn < 300 and < 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labor cost, formulate this as an LPP.

Tailors/Product	$A(x)$	$B(y)$	$Avl$ $
Shirts	$6$	$10$	$60$
Trousers	$4$	$4$	$32$

Let A work x days and B work y days.

$\therefore x \ge 0, y \ge 0$

So L.P.P., objective function Min.

$Z = 300 x + 400 y$

Subject to,

$6x + 10y \ge 60$

$4x + 4y \ge 32$

$\therefore$ LPP

Min.

$Z = 300 x + 400 y$

Subject to,

$3x + 5y \ge 30$

$x + y \ge 8$

$x \ge 0 , y \ge 0$

A fruit grower can use two types of fertilizer in his garden, brand $P$ and brand $Q$ . The amounts (in kg) if nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least $240$ kg of phosphoric acid at least $270$ kg of potash and at most $310$ kg of chlorine.
If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
kg per bag
Brand $P$ Brand $Q$
Nitrogen $3$ $3.5$
Phosphoric acid $1$ $2$
Potash $3$ $1.5$
Chlorine $1.5$ $2$

Let the fruit grower use

$x$ bags of brand

$P$ and

$y$ bags of brand

$Q$ .

The problem can be formulated as follows:

Minimize

$z=3x+3.5y........(1)$

subject tot he constraints

$x+2y\ge 240 .......(2)$

$x+0.5 \ge 90 ......(3)$

$1.5x+2y\le 310 ......(4)$

$x,y \ge 0 ........(5)$

The feasible region determined by the system of constraints is as shown.

The corner points are

$A(240,0), B(140, 50)$ and

$C(20, 140)$

The values of

$z$ at these corner points are as follows.

Corner point

$z=3x+3.5y$

$A(140,50)$

$595$

$B(20.140)$

$550$

$C(40,100)$

$470$

$\rightarrow$ Minimum

The maximum value of

$z$ is

$470$ at

$(40, 100)$

Thus,

$40$ bags of brand

$P$ and

$100$ bags of brand

$Q$ should be added to the garden to minimize the amount of nitrogen.

The minimum amount of nitrogen added to the garden is

$470$ kg.

1033241_423289_ans_177d2bab66d240b683d3c4735ee0b6a0.png

There are two types of fertilizers ${F}_{1}$ and ${F}_{2}$ . ${F}_{1}$ consists of $10$ % nitrogen and $6$ % phosphoric acid and ${F}_{2}$ consists of $5$ % nitrogen and $10$ % phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least $14$ kg of nitrogen and $14$ kg of phosphoric acid for her crop. If ${F}_{1}$ cost Rs. $6$ /kg and ${F}_{2}$ costs Rs. $5$ /kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Let the farmer buy

$x$ kg of fertilizer

${F}_{1}$ and

$y$ kg of fertilizer

${F}_{2}$ . Therefore,

$x\ge 0$ and

$y\ge 0$
The given information can be compiled in a table as follows.

	Nitrogen (%)	Phosphoric Acid (%)	Cost (rs/kg)
${F}_{1}(x)$	$10$	$6$	$6$
${F}_{2}(y)$	$5$	$10$	$5$
Requirement	$14$	$14$

${F}_{1}$ consists of

$10$ % nitrogen and

${F}_{2}$ consists of

$5$ % nitrogen. However, the farmer requires at least

$14$ kg of nitrogen.

$\therefore$

$10$ % of

$x$

$+$

$5$ % of

$y\ge 14$

$\cfrac{x}{10}+\cfrac{y}{20}\ge 14$

$2x+y\ge 280$

${F}_{1}$ consists of

$6$ % phosphoric acid and

${F}_{2}$ consists of

$10$ % phosphoric acid. However, the farmer requires at least

$14$ kg of phosphoric acid

$\therefore$

$6$ % of

$x$

$+$

$10$ % of

$y\ge 14$

$\cfrac{6x}{100}+\cfrac{10y}{100}\ge 14$

$3x+56y\ge 700$
Total cost of fettilizers,

$Z=6x+5y$
The mathematical formulation of the given problem is
Minimise

$Z=6x+5y.....(1)$
subject to the constraints

$2x+y\ge 280......(2)$

$3x+5y\ge 700.....(3)$

$x,y\ge 0......(4)$
The feasible region determined by the system of constraints is as shown
It can be seen that the feasible region is unbounded.
The corner points are

$A(\cfrac{700}{3}),B(100,80)$ and

$C(0,280)$
The values of

$Z$ at these points are as follows

Corner point	$Z=6x+5y$
$A(\cfrac{700}{3},0)$	$1400$
$B(100, 80)$	$1000$	$\rightarrow$ Minimum
$C(0,280)$	$1400$

As the feasible region is unbounded, therefore,

$1000$ may or may ot be the minimum value of

$Z$
For this, we draw a graph of the inequality

$6x+5y< 1000$ and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with

$6x+5y< 1000$
Therefore,

$100$ kg of fertilizer

${F}_{1}$ and

$80$ kg of fertilizer

${F}_{2}$ should be used to minimize the cost. The minimum cost is Rs.

$1000$ .

1038793_423197_ans_bebbef3495be47dd92711c64480eddfe.png

A dietician wishes to mix together two kinds of food $X$ and $Y$ in such a way that the mixture contains at least $10$ units of vitamin $A$ , $12$ units of vitamin $B$ and $8$ units of vitamin $C$ . The vitamin content of one kg of food is given below:
Food Vitamin A Vitamin B Vitamin C
$X$ $1$ $2$ $3$
$Y$ $2$ $2$ $1$
One kg of food $X$ costs Rs. $16$ and one kg of food $Y$ costs Rs. $20$ . Find the least cost of the mixture which will produce the required diet?

Let the mixture contain

$x$ kg of food

$X$ and

$y$ kg of food

$Y$
The mathematical formulation of the given problem is as follows.
Minimize

$z=16x+20y......(1)$

subject to the constraints,

$x+2y\ge 10 ......(2)$

$x+y\ge 6 .........(3)$

$3x+y\ge 8 .......(4)$

$x,y \ge 0..........(5)$

The feasible region determined by the system of constraints is as follows:
The corner points of the feasible region are

$A(10,0), B(2,4), C(1,5)$ and

$D(0,8)$
The values of

$z$ at these corner points are as follows.

Corner point	$z=16x+20y$
$A(10,0)$	$160$
$B(2,4)$	$112$	$\rightarrow$ Minimum
$C(1,5)$	$116$
$D(0,8)$	$160$

As the feasible region is unbounded, therefore,

$112$ may or may not be the minimum value of

$z$
For this, we draw a graph of the inequality,

$16x+20y< 112$ or

$4x+5y< 28$ and check whether the resulting half-plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with

$4x+5y< 28$
Therefore, the minimum value of

$z$ is

$112$ at

$(2,4)$ .

Thus, the mixture should contain

$2$ kg of food

$X$ and

$4$ kg of food

$Y$ . The minimum cost of the mixture is Rs.

$112$

1038803_423247_ans_78c6f35ea1ae4a35b4a12fbaa1adf13a.png

A fruit grower can use two types of fertilizer in his garden, brand $P$ and brand $Q$ . The amounts (in kg) if nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least $240$ kg of phosphoric acid at least $270$ kg of potash and at most $310$ kg of chlorine.
If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
kg per bag
Brand $P$ Brand $Q$
Nitrogen $3$ $3.5$
Phosphoric acid $1$ $2$
Potash $3$ $1.5$
Chlorine $1.5$ $2$

Let the fruit grower use

$x$ bags of brand

$P$ and

$y$ bags of brand

$Q$ .
The problem can be formulated as follows:
Maximize

$z=3x+3.5y........(1)$
subject tot he constraints

$x+2y\ge 240 .......(2)$

$x+0.5 \ge 90 ......(3)$

$1.5x+2y\le 310 ......(4)$

$x,y \ge 0 ........(5)$
The feasible region determined by the system of constraints is as shown.
The corner points are

$A(140,0), B(20, 140)$ and

$C(40,100)$
The values of

$z$ at these corner points are as follows.

Corner point	$z=3x+3.5y$
$A(140,50)$	$595$	$\rightarrow$ Maximum
$B(20.140)$	$550$
$C(40,100)$	$470$

The maximum value of

$z$ is

$595$ at

$(140, 50)$
Thus,

$140$ bags of brand

$P$ and

$50$ bags of brand

$Q$ should be added to the garden to maximize the amount of nitrogen.
The maximum amount of nitrogen added to the garden is

$595$

1039310_423291_ans_40301f8e78954f6684a620d5d143c581.png

A merchant plans to sell two types of personal computers, a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant would stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Let the merchant stock

$x$

$desktop model and$

$y$ $ portable models. Therefore,

$x\ge 0$ and

$y\ge 0$
The cost of a desktop model is Rs.

$25000$ and of a portable model is Rs.

$4000$ . However, the merchant can invest a maximum of Rs.

$70$ lakhs.

$\therefore$

$25000x+40000y\le 7000000$

$5x+8y\le 1400$
The monthly demand of computers will not exceed

$250$ units

$\therefore$

$x+y\le 250$
The profit on a desktop model is Rs.

$4500$ and the profit on a profit model is Rs.

$5000$ .
Total profit,

$Z=4500x+5000y$
Thus, the mathematical formulation of the given problem is
Maximum

$Z=4500x+5000y.......(1)$
subject to the constraints,

$5x+5y\le 1400.........(2)$

$x+y\le 250.......(3)$

$x,y\ge 0...........(4)$
The feasible region determined by the system of constraints is as shown.
The corner points are

$A(250,0), B(200, 50)$ and

$C(0,175)$
The values of

$Z$ at these corner points are as follows

Corner point	$Z=4500x+5000y$
$A(250,0)$	$1125000$
$B(200,50)$	$1150000$	$\rightarrow$ Maximum
$C(0,175)$	$875000$

The maximum value of

$Z$ is

$1150000$ at

$(200,50)$
Thus, the merchant should stock

$200$ desktop models and

$50$ portable models to get the maximum profit of Rs.

$1150000$ .
1038787_423094_ans_929045e122ef48efae167628bf786436.png

A manufacturer makes two types of toys $A$ and $B$ . Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below
Type of toys Machine
$I$ Machine
$II$ Machine
$III$
$A$ $12$ $18$ $6$
$B$ $6$ $0$ $9$
Each machine is available for a maximum of $6$ hours per day. If the profit on each toy of type $A$ is Rs. $7.50$ and that on each toy of type $B$ is Rs. $5$ , show that $15$ toys of type $A$ and $30$ of type $B$ should be manufactured in a day to get maximum profit.

Let

$x$ and

$y$ toys of type

$A$ and

$B$ respectively be manufactured in a day.
The given problem can be formulated as follows:
Maximise

$z=7.5x+5y......(1)$
subject to the constraints

$2x+y\le 60........(2)$

$x\le 20 ........(3)$

$2x+3y\le 120.........(4)$

$x,y\ge 0..........(5)$
The feasible region determined by the constraints is as shown
The corner points of the feasible region are

$A(20,0), B(20,20), C(15,30)$ and

$D(0,40)$
The value of

$z$ at these corner points are as follows.

Corner point	$Z=7.5x+5y$
$A(20,0)$	$150$
$B(20,20)$	$150$
$C(15,30)$	$262.5$	$\rightarrow$ Maximum
$O(0,40)$	$200$

The maximum value of

$z$ is

$262.5$ at

$(15,30)$
Thus, the manufacturer should manufacture

$15$ toys of type

$A$ and

$30$ toys of type

$B$ to maximize the profit.
1038820_423250_ans_0634336b388047f6af79ba1af62418f0.png

The graph of 3x + 2y = 6 meets the x-axis at point P and the y-axis at point Q. Use the graphical method,then the co-ordinates of points P and Q are P = (2, 0) and Q = (0, 3)
If true then enter $1$ and if false then enter $0$

To draw the graph for the given linear equation, we need to first plot

the points which lie on it, join them to get a straight line.

First equation is

$x+y = 2 => y = -x + 2$
Substitute different values of x in the equation and calculate to get corresponding values of y.

$x$	$y= -x + 2$	Point $(x,y)$
$0$	$2$	$(0,2)$
$2$	$0$	$(2,0)$
$4$	$-2$	$(4,-2)$
$-1$	$3$	$(-1,3)$
$-3$	$5$	$(-3,5)$

Plot these points on the graph paper and join them using a straight line.

Second equation is

$x -2y = 5 => y = \frac {x}{2} - \frac {5}{2}$
Substitute different values of x in the equation and calculate to get corresponding values of y.

$x$	$y=\frac {x}{2} - \frac {5}{2}$	Point $(x,y)$
$0$	$- \frac {5}{2}$	$(0,- \frac {5}{2})$
$3$	$-1$	$(3,-1)$
$5$	$0$	$(5,0)$
$-2$	$- \frac {7}{2}$	$(-2, - \frac {7}{2})$
$-3$	$-4$	$(-3,-4)$

Plot these points on the graph paper and join them using a straight line.

Third equation is

$\frac {x}{3} + y = 0 => y = - \frac {x}{3}$
Substitute different values of x in the equation and calculate to get corresponding values of y.

$x$	$y= - \frac {x}{3}$	Point $(x,y)$
$0$	$0$	$(0,0)$
$3$	$-1$	$(3, -1)$
$6$	$-2$	$(6, -2)$
$-3$	$1$	$(-3,1)$
$-6$	$2$	$(-6,2)$

Plot these points on the graph paper and join them using a straight line.

All the three lines meet at a point.

$(3,-1)$

An oil company has two depots $A$ and $B$ with capacities of $7000L$ and $4000L$ respectively. The company is to supply oil to three petrol pumps $D, E$ and $F$ whose requirements are $4500L, 3000L$ and $3500L$ respectively. The distance (in km) between the depots and the petrol pumps is given in the following table:
Distance (in km)
From/To $A$ $B$
$D$ $7$ $3$
$E$ $6$ $4$
$F$ $3$ $2$
Assuming that the transportation cost of $10$ litres of oil is Re. $1$ per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Let

$x$ and

$y$ litres of oil be supplied from

$A$ to the petrol pumps,

$D$ and

$E$ . Then

$(7000-x-y)$ will be supplied from

$A$ to petrol pump

$F$ .
The requirement at petrol pum[

$D$ is

$4500L$ . Since

$x$

$L$ are transported from depot

$A$ , the remaining

$(45000-x)$

$L$ will be transported from petrol pump

$B$ $.
Similarly,

$(3000-y)L$ and

$3500(7000-x-y)=(x+y-3500)L$ will be transported from depot

$B$ to petrol pump

$E$ and

$F$ respectively.
The given problem can be represented diagrammatically as shown in first figure.

$x\ge 0, y\ge 0$ and

$(7000-x-y)\ge 0$

$\Rightarrow$

$x\ge, y\ge 0$ and

$x+y\le 7000$

$4500-x\ge 0, 3000-y\ge 0$ and

$x+y-3500\ge 0$

$\Rightarrow$

$x\le 4500, y\le 3000$ and

$x+y\ge 3500$
Cost of transporting

$10L$ of petrol

$=$ Re.

$1$
Cost of transporting

$1L$ of petrol

$=$ Rs.

$\cfrac{1}{10}$
Therefore, total transportation cost is given by

$z=\cfrac{7}{10}\times x+\cfrac{6}{10}\times y+\cfrac{3}{10}(7000-x-y)+\cfrac{3}{10}(4500-x)+\cfrac{4}{10}(3000-y)+\cfrac{2}{10}(x+y-3500)$

$=0.3x+0.1y+3950$
The problem can be formulated as follows:
Minimize

$z=0.3x+0.1y+3950 ......(1)$
subject to the constraints

$x+y\ge 7000 ......(2)$

$x\le 4500 ........(3)$

$y\le 3000 ........(4)$

$x+y\ge 3500 ....(5)$

$x,y \ge 0........(6)$
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are

$A(3500, 0), B(4500, 0), C(4500, 2500), D(4000,3000)$ and

$E(500, 3000)$
The values of

$z$ at these corner points are as follows

Corner point	$z=0.3x+0.1y+3950$
$A(3500,0)$	$5000$
$B(4500,0)$	$5300$

$C(4500, 2500)$	$5550$
$D(4000, 3000)$	$5450$
$E(500,3000)$	$4400$	$\rightarrow$ Minimum

The minimum value of

$z$ is

$4400$ at

$(500, 3000)$
Thus, the oil supplied from depot

$A$ is

$500L, 3000L$ and

$3500L$ and from depot

$B$ is

$4000L, 0L$ and

$0L$ to petrol pumps

$D, E$ and

$F$ respectively.
The minimum transportation cost is Rs.

$4400$

1038825_423288_ans_938fc135b67e464bbd7609599004adc7.png

A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires $9$ labour hours of fabricating and $1$ labour hour for finishing. Each type of B requires $12$ labour hours for fabricating and $3$ labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are $180$ and $30$ respectively. The company makes a profit of $80$ on each piece of type A and $120$ on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

Let the pieces of type A manufactured per week be

$x$

Let the pieces of type B manufactured per week be

$y$

Therefore maximum profit

$z=80x+120y$

Fabricating hours for

$A$ is

$9$ and finishing hour is

$1$

Fabricating hours forours

$B$ is

$12$ and finishing hours is

$3$

Maximum number of fabricating hours =

$180$

$\therefore 9x+12y\le 180\\ i.e^3x+4y\le 60$

Maximum number of finishing hours=

$30$

$x+3y\le 30$

Now

$z$ is

$=80+120y$ is subject to

$3x+4y\le 60$

$x+3y\le 30$

$y\ge 0$

Now the area of th feasible region is as shown in above figure

$3x+4y=60$

$3x+3y=30$

$3x+4y=60$

$3x+9y=30$

$\therefore y=6$

and

$x=12$

W can see that the post on the region are

$A(0,15) , B(0,10) , C(20,0)$ and

$D(12,6)$

Th maximum profit is

Point(x,y)	$z=80x+120$
A(0,15)	$1800\rightarrow$ maximum
B(0, 10)	1200
C(20, 0)	1600
D(12,6)	1680

Hence the maximum profit is at

$(0,15)$

$\therefore$ The teaching aid

$A=0$ and teaching aid

$B=15$ has to be made.

You need to buy some filing cabinets. You know that Cabinet $X$ costs $$10$ per unit, requires six square feet of floor space, and holds eight cubic feet of files. Cabinet $Y$ costs $$20$ per unit, requires eight square feet of floor space, and holds twelve cubic feet of files. You have been given $$140$ for this purchase, though you don't have to spend that much. The office has room for no more than $72$ square feet of cabinets. How many of which model should you buy, in order to maximise storage volume. Write the maximum storage volume in the provided box.

Let number of X models be

$x$

Let number of Y models be

$y$

For floor

$6x+8y=72$

$3x+4y=36$

$10x+20y\le 140$

Volume

$=8x+12y$

Now

$x+2y\le 14$

Hence

$2x+4y\le 28$

Volume

$=8x+12y=8x+108-9x=108-x$

$2x+4y\le 28$

$2x+36-3x\le 28$

$36-x\le 28$

$x\ge 8$

So volume is maximum when

$x$ is minimum

$=8$

So volume

$=108-8=100$

A dietician wishes to mix two types of foods in such away that the vitamin contents of the mixture contains at least $8$ units of vitamin A and $10$ units of vitamin C. Food I contains $2$ units/kg of vitamin A and $1$ unit/kg of vitamin C while Food II contains $1$ unit/kg of vitamin A and $2$ units/kg of vitamin $1$ unit/kg of vitamin C. It costs $Rs. 5$ per kg to purchase food I and $Rs. 7$ per kg to purchase Food II. Determine the maximum cost of such a mixture. Formulate the above as a LPP and solve it graphically.

Suppose the dietician mixes

$x$ kg of food

$I$ and

$y$ kg of food

$II.$

The first condition of the mixture containing

$8$ units of vitamin A is given by

$2x + y \geq 8$

Also, atleast

$10$ units of vitamin C is given by

$x + 2y \geq 10$

The total cost of the mixture is given by

$5x + 7y$

So, our problem becomes to maximise

$5x + 7y$ , subject to the conditions

$2x + y \geq 8 \ \ \ ...(1)$ and

$x + 2y \geq 10 \ \ \ ...(2)$

Equation (1) multiplied by

$2$ gives

$4x + 2y \geq 16$ and when subtracted with equation (2), we get

$3x \geq 6$ or

$x \geq 2$

When substituted in equation (1), we obtain

$y \geq 4$

The minimum cost of such a mixture thus becomes

$10 + 28 = Rs. 38. \\$

However, the maximum cost can go upto infinity.

Formation of LPP

To Maximize:	$5x + 7y$
Constraints:	$2x + y \geq 8$
	$x + 2y \geq 10$
	$x≥0, y≥0$

Verifying values at corner points:

Corner Points	$(0,8)$	$(2,4)$	$(10,0)$
Value of $5x + 7y$	$56$	$38$	$50$

Minimum at

$(2,4)$ , and It can attain infinity as it's maximum value.

A toy company manufactures two types of dolls, $A$ and $B$ . Market tests and available resources have indicated that the combined production level should not exceed $1200$ dolls per week and the demand for dolls of type $B$ is at most half of that for dolls of type $A$ . Further, the production level of dolls of type $A$ can exceed three times the production of dolls of other type by at most $600$ units. If the company makes profit of $Rs.\ 12$ and $Rs.\ 16$ per doll respectively on dolls $A$ and $B$ , how many of each should be produced weekly in order to maximize the profit?

Let

$x$ and

$y$ be the number of dolls of type

$A$ and

$B$ respectively that are produced per week
The given problem can be formulated as follows
Maximize

$z=12x+16y........(1)$
subject to the constraints

$x+y\le 1200 .........(2)$

$y\le \cfrac{x}{2}$

$\Rightarrow$

$x\ge 2y ......(3)$

$x-3y\le 600 ........(4)$

$x,y\ge 0 ........(5)$
The feasible region determined by the system of constraints is as shown.
The corner points are

$A(600,0), B(1050,150)$ and

$C(800,400)$
The values of

$z$ at these corner points are as follows

Corner point	$z=12x+16y$
$A(600,0)$	$7200$
$B(1050,150)$	$15000$
$C(800,400)$	$16000$	$\rightarrow$ Maximum

The maximum value of

$z$ is

$16000$ at

$(800,400)$
Thus,

$800$ and

$400$ dolls of type

$A$ and type

$B$ should be produced respectively to get the maximum profit of Rs.

$16000$

1039355_423424_ans_50728f3b1b4d46c5bc609772636ba35c.png

A cooperative society of farmers has $50$ hectares of land to grow two crops A and B. The profits from crops A and B per hectare are estimated as Rs $10,500$ and Rs $9,000$ respectively. To control weeds, a liquid herbicide has to be used for crops A and B at the rate of $20$ litres and $10$ litres per hectare, respectively. Further not more than $800$ litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. Keeping in mind that the protection of fish and other wildlife is more important than earning profit, how much land should be allocated to each crop so as to maximize the total profit? Form an LPP from the above and solve it graphically. Do you agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment?

Let the land allocated to crop

$X$ be

$x$ hector and land allocated to crop

$Y$ be

$y$ hector.

Given: Herbicides used for crop

$X=20$ litres per hectare

Herbicides used for crop

$Y=10$ litres per hectare

Maximum quantity of herbicide=

$800$ litres

$\therefore 20x+10y\le 800$

$2x+y<800$

The total land available =

$50$ hectare

$\therefore x+y\le 50$

As we count to maximise the profit

Hence, the function used here is Maximise z

Profit from Crop

$x=$ Rs.

$10500$ per hectare

Profit from Crop

$y=$ Rs.

$9000$ per hectare

Maximize

$z:10500x+9000y$

Combining all constraints

$2xy\le 80$

$x+y\le 50$

$x,y\ge 0$

$x+y\le 50$

$x$	0	50
$y$	50	0

ii)

$2x+y\le 80$

$x$	0	40
$y$	80	0

Corner Points	Value of z
(0,50)	450000
(30,20)	495000 $\rightarrow$ maximum
(40,0)	420000
(0,0)	0

Hence the profit will be maximum of

Crop

$x=30$ hectare

Crop

$y=20$ hectare

Maximum profit

$=$ Rs.

$4,95,000$

A cooperative society of farmers has 50 hectares of land to grow two crops A and B. The profits from crops A and B per heclare are estimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops A and B at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. Keeping in mind that the protection of fish other wildlife is more important than earning profit, how much land should be allocated to each crop so as to maximize the total profit?
From an LPP from the above and solve it graphically. Do you agree with the massage that the protection of wildlife is utmost necessary to preserve the balance in environment?

Let

$x$ hectares fro crop

$A$ and

$y$ hectares for crop

$B$ be allocated.

$\textbf{A}$	$\textbf{B}$
$x$	$y$	$50$ hectares
$10500$	$9000$	Profit
$20 \ L/H$	$10\ L/H$	$800$ litres atmost

We need to maximize profit given by

$Z=10500x+9000y$ subject to

$x+y\leq 50$

$20x+10y\leq800$

$x,y\geq 0$

After plotting graph, the corner points are

Corner Point	Value of $Z$
$O \,(0,0)$ $A \,(40, 0)$ $B\, (30, 20)$ $C\, (0, 50)$	$Z=0$ $Z= 420000$ $Z = 315000+180000=495000$ $Z= 450000$	Maximum

Therefore, $30$ hectares for crop $A$ and $20$ hectares for crop $B$ .

A manufacturing company makes two types of teaching aids $A$ and $B$ of Mathematics for Class XII. Each type of $A$ requires $9$ labour hours of fabricating and $1$ labour hour for finishing. Each type of $B$ requires $12$ labour hours for fabricating and 3 labour hours for finishing, the maximum labour hours available per week are $180$ and $30$ respectively. The company makes a profit of $Rs. 80$ on each piece of type $A$ and $Rs. 120$ on each piece of type $B$ . How many pieces of type $B$ should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

${\textbf{Step-1: Assume pieces per week and find equations for maximum and minimum time. }}$

${\text{Let the pieces of type A manufactured per week be}}$ $x.$

${\text{Let the pieces of type B manufactured per week be}}$ $y.$

${\text{Maximum profit}}$ $z = 80x + 120y$

${\text{Fabricating hours for A is 9 and finishing hours is 1}}$

${\text{Fabricating hours for B is 12 and finishing hours is 3}}$

${\text{Maximum number of fabricating hours is 180.}}$

$9x + 12y$ $\leqslant$ $180$

$3x + 4y$ $\leqslant$ $60$

${\text{Maximum number of finishing hours is 30.}}$

$x + 3y$ $\leqslant$ $30$

${\textbf{Step-2: Solve the equations.}}$

${\text{Thus we can formulate the linear programming problem as follows:}}$

${\text{Max}}$ $z = 80x + 120y$ ${\text{subjected to the constraints,}}$ $3x + 4y$ $\leqslant$ $\leqslant$ $30,x$ $\geqslant$ $\geqslant$ $0$

${\text{Solving the constraints we get,}}$ ,

$x = 12, y = 6.$

${\text{Now, the area of feasible region is as shown in the figure.}}$

${\text{We can see that the points bounded by the feasible region are}}$ $A(0,10),B(12,6),C(20,0)$

${\textbf{Step-3: Calculate the maximum profit.}}$

$At A(0,10), z = 80(0) + 120(10) = 1200$

$At B(12,6), z = 80(12) + 120(6) = 1680$

$At C(20,0), z = 80(20) + 120(0) = 1600$

${\text{Hence the maximum profit is at (12, 6)}}$

${\textbf{Therefore 12 pieces of type A and 6 pieces of type B should be manufactured}}$ ${\textbf{per week to get a maximum profit of Rs.1680.}}$

2168085_505005_ans_6e1c9e50aa4f4e0196edf7cd622156ca.jpeg

A dietician wishes to mix two kinds of food $X$ and $Y$ in such a way that the mixture contains at least $10$ units of vitamin $A, 12$ units of vitamin $B$ and $8$ units of vitamin $C$ . The vitamin contains of per kg food is given below.
Food Vitamin $A$ Vitamin $B$ Vitamin $C$
$X$ $1$ unit $2$ units $3$ units
$Y$ $2$ units $2$ units $1$ unit
One kg of food $X$ costs rupees $24$ and one kg of food $Y$ costs rupees $36$ . Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.

Let

$x$ be the number of units of food X and

$y$ be the number of units of food Y be mixed to obtain the desired diet. Then LPP of the given problem is :
Minimise,

$z=24x+36y$
Subject to the constraints,

$x+2y \ge 10, 2x+2y\ge 12$ i.e.,

$x+y\ge 6$

$3x+y\ge 8$

$x\ge 0, y\ge 0$
We draw the lines

$x+2y=10, x+y=6, 3x+y=8$ and obtain the feasible region (unbounded and convex) shown in the figure.
Thus corner points are

$A(0, 8), B(1, 5), C(2, 4)$ and

$D(10, 0)$ .
The values of

$z$ (in rupees) at these points are given in the following table :

Corner Point	Objective Function $z=24x+36y$
$A(0, 8)$	$z=24\times 0+36\times 8=288$
$B(1, 5)$	$z=24\times 1+36\times 5=204$
$C(2, 4)$	$z=24\times 2+36\times 4=192$
$D(10, 0)$	$z=24\times 10+36\times 0=240$

As the feasible region is unbounded, we draw the graph of the half plane

$6x+10y<52$ i.e.,

$3x+5y<26$ and note that there is no point common with the feasible region.
Therefore,

$z$ has the minimum value of the minimum value is Rupees

$192$ .
It occurs at

$C(2, 4)$ .
i.e., when

$2\ kg$ of food

$X$ and

$4\ kg$ of food

$Y$ are mixed to get the desired diet.

A dealer in rual area wishes to purchase a number of sewing machines. He has only $Rs\ 5,760$ to invest and has space for at most $20$ items for storage. An electronic sewing machine costs him $Rs\ 360$ and a manually operated sewing machine $Rs\ 240$ . He can sell an electronic sewing machine at a profit of $Rs\ 22$ and a manually operated sewing machine at a profit of $Rs\ 18$ . Assuming that he can sell all the items that he can buy how should he invest his money in order to maximize his profit? Formulate this problem mathematically and then solve it.

Let

$x$ and

$y$ be electronic and manually operated sewing machines purchased respectively.
Therefore, LPP is maximize

$P = 22x + 18y$
Subjected to

$\left.\begin{matrix}360x+240y\leq5760\\or, 3x+2y\leq 48 \\ x + y\leq 20\\x\geq0, y\geq 0 \end{matrix}\right\}$

For correct graph vertices of feasible region are

$A(0, 20), B(8, 12), C(16, 0) and O(0, 0)$

$P(A) = 360, P(B) = 392, P(C) = 352$

$3x + 2y = 48-- (1)$

$x + y = 20--- (2)$

Solving equation

$(1)$ and

$(2),$ we get

$x = 8, y = 12$

Therefore for maximum

$P,$ electronic machines

$= 8.$

Manual machines

$= 12.$

A company manufactures three kinds of calculators : A, B and C in its two factories I and II, The company has got an order for manufacturing at least $6400$ calculators of kind A, $4000$ of kind B and $4800$ of kind C. The daily output of factory I is of $50$ calculators of kind A, $50$ calculators of kind B, and $30$ calculators of kind C. The daily output of factory II is of $40$ calculators of kind A, $20$ of kind B and $40$ of kind C. The cost per day to run factory I is Rs. $12,000$ and of factory II is Rs. $15,000$ .
How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this problem as an LPP and solve it graphically.

Let factories I and II should be operated for x and y number of days respectively. Then the problem can be formulated as in L.P.P. as:

Minimise

$Z = 12000x + 15000y$
Subject to constraints

$50x+40y\geq6400$ i.e.,

$5x+4y\geq 640$

$50x+20y\geq4000$ i.e.,

$5x+2y\geq400$

$30x+40y\geq4800$ i.e.,

$3x+4y\geq480$

$x\geq0, y\geq 0$

We draw the lines

$5x+4y\geq 640, 5x+2y\geq400, 3x+4y\geq480$ and obtain the feasible region (unbounded and convex) shown shaded in the adjoining figure. the corner points are

$A(0, 200), B(32, 120), C(80, 60)$ and

$D(160, 0)$ .

The values of Z at these points are

$3000000, 2184000, 1860000$ and

$1920000$ respectively. As the feasible region is unbounded, we draw the graph of the half plane.

$12000x+15000y<1860000$
i.e.,

$12x + 15y < 1860$ and note that there is no point common with the feasible region, therefore Z has minimum and minimum value is Rs.

$1860000$ .

It occurs at the point

$(80, 60)$ i.e., Factory I should be operated for

$80$ days and factory II should be operated for

$60$ days to minimise the cost.

Running Cost per day (in Rs )		$A$	$B$	$C$
$12000$	$I$	$50$	$50$	$30$
$15000$	$II$	$40$	$20$	$40$

A company manufactures two types of products $A$ and $B$ . Each unit of $A$ requires $3$ grams of nickel and $1$ gram of chromium, while each unit of $B$ requires $1$ gram of nickel and $2$ grams of chromium. The firm can produce $9$ grams of nickel and $8$ grams of chromium. The profit is Rs. $40$ on each unit of product of type $A$ and Rs. $50$ on each unit of type $B$ . How many units of each type should the company manufacture so as to earn maximum profit? Use linear programming to find the solution

Type of product	$9$ gms Nickel (gms)	$8$ gms Chromium (gms)	Profit per unit
$A$	$3$	$1$	Rs. $40$
$B$	$1$	$2$	Rs. $50$

Let

$x=$ number of units of type

$A$

$y=$ Number of units of type

$B$
Maximize

$Z=40x+50y$
Subject to the constraints

$3x+y\le 9$

$x+2y\le 8$
and

$x,y\ge 0$
Consider the equation,

$3x+y=9$

$x$	$=$	$3$
$y$	$=$	$0$

and

$x+2y=8$

$x$	$=$	$8$
$y$	$=$	$0$

The solution set of this system is the shaded region in the diagram
Now, we can determine the maximum value of

$Z$ by evaluating the value of

$Z$ at the four points (vertices) is shown below

Vertices	$Z=40x+50y$
$(0,0)$	$Z=40\times 0+50\times 0=$ Rs. $0$
$(3,0)$	$Z=40\times 3+50\times 0=$ Rs. $120$
$(0,4)$	$Z=40\times 0+50\times 4=Rs.200$
$(2,3)$	$Z=40\times 2+50\times 3=Rs.230$

From graph,
Maximum profit,

$Z=$ Rs.

$230$

$\therefore$ Number of units of type

$A$ is

$2$ and number of units of type

$B$ is

$3$ .

Find graphically, the maximum value of $z=2x+5y$ , subject to constraints given below:
$2x+4y\le 8$
$3x+y\le 6$
$x+y\le 4$
$x\ge 0, y \le 0.6$

Maximise

$z = 2x + 5y$ , subject to the constraints

$2x+4y\leq 8\Rightarrow x+2y\leq 4$

$3x+y\leq 6, x + y\leq 4, x\geq 0, y\geq 0$ .

Draw the lines

$x + 2y = 4$ (passes through

$(4, 0), (0, 2)$ );

$3x + y = 6$ (passes through

$(2, 0), (0, 6)$ ) and

$x + y = 4$ (passes through

$(4, 0), (0, 4)$ ).
Shade the region satisfied by the given inequations.

The shaded region in the figure gives the feasible region determined by the given inequations.

Solving

$3x + y = 6$ and

$x + 2y = 4$ simultaneously, we get

$x=\dfrac{8}{5}$ and

$y=\dfrac{6}{5}$

We observe that the feasible region OABC is a convex polygon and bounded and has corner points.

$O(0, 0), A(2, 0), B(\dfrac{8}{5}, \dfrac{6}{5}), C(0, 2)$

The optimal solution occurs at one of the corner points.
At

$O(0, 0), z = 2.0+5.0=0;$
At

$A(2, 0), z = 2.2+5.0=4;$

$B(\dfrac{8}{5}, \dfrac{6}{5}),z = 2.\dfrac{8}{5}+5.\dfrac{6}{5}=\dfrac{46}{5};$
At

$C(0, 2), z = 2.0+5.2=10;$

Therefore,

$z$ maximum value at

$C$ and maximum value

$= 10.$

Minimize: $Z=6x+4y$
Subject to : $3x+2y \ge 12$ ,
$x+y\ge 5$ ,
$0\le x\le 5$ ,
$0\le y\le 6$

The feasible region is

$DEAY$ which is shaded in the figure.
The vertices of the feasible region are

$A(0,6)$ and

$D(5,0)$ .

$E$ is the point of intersection of the lines. Now solving equation

$(i)$ and

$(ii)$ we get intersecting point

$E(2,3)$ . Origin do not satisfied given condition

$(0,0)$ does not present in under lined area.
Calculation of minimum value:

Corner points	$Z=6x+4y$
At $A(0,6)$	$Z=6\times 0+4\times 6=24$
At $E(2,3)$	$Z=6\times 2+4\times 3=24$
At $D(5,0)$	$Z=6\times 5+4\times 0=30$

Since

$Z$ has minimum value

$24$ at two consecutive vertices

$A$ and

$E$ of the feasible region,

$Z$ is minimum value

$24$ at every point of segment

$AE$ . Hence, there are infinite number of optimal solutions.

A mill owner buys two types of machines $A$ and $B$ for his mill. Machine $A$ occupies $1,000$ sqm of area and requires $12$ men to operate it; while machine $B$ occupies $1,200$ sqm of area and requires $8$ men to operate it. The owner has $7,600$ sqm of area available and $72$ men to operate the machines. If machine $A$ produces $50$ units and machine $B$ produces $40$ units daily, how many machines of each type should he buy to maximize the daily output? Use Linear Programming to find the solution.

The data given in the problem are as under:

	Machine A	Machine B	Maximum available
Area needed	$1,000$ sq. m	$1,200$ sq.m	$7,600$ sq.m
Labour force	$12$	$8$	$72$
Daily output	$50$ units	$40$ units	-

Let

$x$ and

$y$ be the number of machines A and B respectively

$\left.\begin{matrix}1,000 x + 1,200 y \leq 7,600\\ 10x + 12y \leq 76\\ 5x + 6y \leq 38\\ 12x + 8y \leq 72\\ 3x + 2y \leq 18\\ x \geq 0, y \geq 0\end{matrix}\right\} constraints$

Total output

$Z = 50x + 40y$

The problem is to maximise

$Z = 50x + 40y$ subject to constraints

$5x + 6y \leq 38$

$3x + 2y \leq 18$

$x \geq 0, y \geq 0$

$3x + 2y = 18$

$x$	$6$	$0$	$4$
$y$	$0$	$9$	$3$

$5x + 6y = 38$

$x$	$\displaystyle \frac{38}{5}$	$0$	$4$
$y$	$0$	$\displaystyle \frac{19}{3}$	$3$

The vertices of the feasible region ORPD are

$O(0, 0), R(6, 0), P(4, 3)$ and

$D\left (0, \dfrac{19}{3}\right )$

Point	Value of $Z=50x+40y$
At $O(0,0)$	$Z = 0$
At $R(6, 0)$	$Z=50\times 6+0=300$
At $P(4, 3)$	$Z=50\times 4+40\times 3=320$
At $D\left (0, \dfrac{19}{3}\right )$	$Z=50\times 0+40\times \frac{19}{3}$ $=\frac{760}{3} =2.53.33$

Thus we see that

$Z$ is maximum at

$(4, 3)$ .

$\therefore$ Number of machine

$A = 4$
Number of machine

$B = 3$
The graph of these in equation is as shown

Solve the following L.P.P. graphically
Maximize: $z = 10x + 25y$
Subject to: $x\leq 3 y\leq 3, x + y \leq 5, x \geq 0, y \geq 0$

We have,

$z = 10x + 25y$ .... (i)
The constraints are

$x \leq 3$ ....(ii)

$y \leq 3$ .... (iii)

$x + y \leq 5$ .... (iv)

$x \geq 0$ .... (v)

$y \geq 0$ ....(vi)
The constraints (ii) and (iii) represent the closed half-planes on the left of the line

$x = 3$ and below

$y = 3$ respectively.
The line corresponding to (iv) is

$x + y = 5$ ...(vii)

$(5, 0)$ and

$(0, 5)$ lie on (vii). The origin does not lie on (vii) and it lies in half-plane of (vii) of

$0 + 0 \leq 5$ , which is true.

$\therefore$ The closed half plane containing the origin in the graph of (iv).
The constraints (v) and (vi) represent the closed half-planes on the right of

$y$ -axis and above

$x$ -axis respectively.
The shaded bounded region is the feasible region of the given L.P.P. we use curve point method to maximize

$z$ . The vertices of the feasible region are

$O(0, 0), A[0, 3), B(2, 3), C(3, 2)$ and

$D(3, 0)$
At

$O(0, 0) \ z = 10(0) + 25(0) = 0$
At

$A(0, 3) \ z = 10(0) + 25(3) = 75$
At

$B(2, 3) \ z = 10(2) + 25(3) = 95$
At

$C(3, 2) \ z = 10(3) + 25(2) = 80$
At

$D(3, 0) \ z = 10(3) + 25(0) = 30$

$\therefore$ maximum value of

$z = 95$ , where

$x = 2$ and

$y = 3$ .

Two tailors P and Q eart Rs. 150 and Rs. 200 per day respectively. P can stitch 6 shirts and 4 trousers a day, while Q can stitch 10 shirts and 4 trousers per day. How many days should each work to produce at least 60 shirts and 32 trousers at minimum labour cost?

Let the tailor P work for x days and the tailor Q work for y days respectively.

$Z = 150 x + 200 y$
Here the problem can be formulated as an L.P.P. as follows
Minimize

$Z = 150x + 200y$
Subject to the constraints

$6x + 10 y \geq 60$

$\Rightarrow 3x + 5y \geq 30$

$4x + 4y \geq 32$

$x + y \geq 8$
and

$x \geq 0, y \geq 0$

Converting them into equations we obtain the following equations

$3x + 5y = 30$ ,

$\Rightarrow y = \dfrac{30 - 3x}{5}$

$x$	$0$	$10$	$5$
$y$	$6$	$0$	$3$

$\Rightarrow x + y = 8$

$y = 8 -x$

$x$	$0$	$8$	$5$
$y$	$8$	$0$	$3$

The lines are shown on the graph paper and the feasible region (Unbounded convex) is shown shaded.
The corner points are

$A(10, 0), B(5, 3)$ and

$C(0, 8)$
At the corner point the value of

$Z = 150x + 200y$
At

$A (10, 0) Z = 1500$
At B(5, 3),

$Z= 150 \times 5 + 200 \times 3 = 750 + 600 = 1350$
At C (0, 8),

$Z = 1600$
As the feasible region is unbounded, we draw the graph of the half-plane

$150x + 200y < 1350$

$3x + 4y < 27$
There is no point common with the feasible region, therefore, Z has minimum value.
Minimum value of

$Z$ is Rs.

$1350$ and it occurs at the point

$B(5, 3)$ .
Hence, the labour cost is Rs. 1350 when P works for 5 days and Q works for 3 days.

A diet of a sick person must contains at least $48$ units of vitamin $A$ and $64$ units of vitamin $B$ . Two foods $F_1$ and $F_2$ are available. Food $F_1$ costs Rs. $6$ per unit and Food $F_2$ costs Rs. $10$ per unit. One unit of food $F_1$ contains $6$ units of vitamin $A$ and $7$ units of vitamin $B$ . One unit of food $F_2$ contains $8$ units of vitamin $A$ and $12$ units of vitamin $B$ . Find the minimum cost for the diet that consists of mixture of these two foods and also meeting the minimum nutritional requirements.

Given, Vitamin

$A \geq 48$ units and Vitamin

$B \geq 64$ units.

$\therefore F_1 = 6A + 7B$ and

$F_2 = 8A + 12B$
Let the required number of units for

$F_1$ be

$x$ and for

$F_2$ be

$y$ .
Thus

$x F_1 + yF_2 \geq 48A + 64 B$

$\Rightarrow x(6A + 7B) + y(8A + 12 B) \geq 48 A + 64 B$

$\Rightarrow A(6x + 8y) + B (7x + 12 y) \geq 48A + 64 B$
By comparing, we get

$6x + 8y \geq 48$

$3x + 4y \geq 24$ ......... (i)
and

$7x + 12 y \geq 64$ ........ (ii)
To find the intersection, consider equality. Multiplying equation (i) by

$3$ and subtracting equation (ii) from it,

$9x + 12 y = 72$

$7x + 12 y = 64$
Thus

$2x =8$

$\Rightarrow x = 4$
Thus

$3(4) + 4y = 24$

$\Rightarrow 12 + 4y = 24$

$\Rightarrow 4y = 12$

$\Rightarrow y = 3$
Also,

$F_1 = Rs. 6$ and

$F_2 = Rs. 10$
Hence,

$4F_1 + 3F_2 = Rs [4(6) + 3(10)]$

$= Rs [24 + 30] = Rs. 54$

Solve the following linear programming problem graphically:
Maximize: $Z=60x+40y$
subject to the constraints:
$x+2y\le 12$ ; $2x+y\le 12$
$x+\cfrac{5}{4}y\ge 5;x\ge 0, y\ge 0$ .

Function to maximize:	$Z = 60x + 40y$
Constraints:	$x+2y≤12$
	$2x+y≤12$
	$x+\cfrac54y≥5$
	$x \geq 0, y\geq 0$

Cornor Points	Values of $Z = 60x+40y$ at cornor points
${(0, 4)}$	$160$
$(0,6)$	$240$
$\textbf{(4, 4)}$	$\textbf{400}\ \ \leftarrow$ Maximum
$(6,0)$	$360$
$(5,0)$	$300$

Hence, the maximum value of

$Z = 60x+40y$ is

$400$ .

A company manufactures two types of toys $A$ and $B$ . A toy of type $A$ requires $5$ minutes for cutting and $10$ minutes for assembling. A toy of type $B$ requires $8$ minutes for cutting and $8$ minutes for assembling. There are $3$ hours available for cutting and $4$ hours available for assembling the toy in a day. The profit is $Rs. 50$ each on a toy of type $A$ and $Rs. 60$ each on a toy of type $B$ . How many toys of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution

	Toy $A$ $\quad$	Toy $B$ $\quad$	Time in a day
Cutting time	$5\ min$	$8\ min$	$180\ min$
Assembling time	$10\ min$	$8\ min$	$240\ min$
Profit	$50$	$60$
Assumed quantity	$x$	$y$

Profit function

$z = 50x + 60y$

$5x + 8y = 180$

	$A$	$B$
$x$	$0$	$36$
$y$	$22.5$	$0$

$x \geq 0, y \geq 0$

$5x + 8y \leq 180$

$10x + 8y \leq 240$ or

$5x + 4y \leq 120$

$5x + 4y = 120$

	$C$	$D$
$x$	$0$	$24$
$y$	$30$	$0$

Corner point	$z = 50x + 60y$
At $O(0, 0)$	$0$
At $D (24, 0)$	$1200$
At $E(12, 15)$	$1500$
At $A(0, 22.5)$	$1350$

Hence the maximum profit is

$Rs. 1500$ at

$E(12, 15)$ .

Solve the following linear programming problem by graphical method:
Maximize $Z = 3x + 2y$
subject to the in constraints
$x + 2y \leq 10, 3x + y \leq 15, x, y\geq 0$

Function to Maximize:	$Z = 3x + 2y$
Constraints:	$x + 2y \geq 10$
	$3x + y \leq 15$
	$x \geq 0, y\geq 0$

Cornor Points	Values of $Z = 3x+2y$ at cornor points
${(0, 0)}$	$0$
$(0,5)$	$10$
$\textbf{(4, 3)}$	$\textbf{18}\ \ \ \leftarrow$ Maximum
$(5,0)$	$15$

Hence, the Maximum value of

$Z = 3x+2y$ is

$18$ .

An aeroplane can carry a maximum load of $200$ passengers. Baggage allowed to the first class ticket holder is $30$ kg and for the economy class ticket holder is $20$ kg. Maximum capacity of the aeroplane to carry the baggage is $4500$ kg. The profit on each first class ticket is Rs. $500$ and on each economy class ticket is Rs. $300$ . Formulate the problem, as L.P.P to maximize the profit.

Let

$x$ and

$y$ be the number of first class ticket holders and economy class ticket holders respectively.

$x\ge 0$

$y\ge 0$

$x+y\le 200$

$30x+20y\le 4500$

$\Rightarrow$

$3x+2y\le 450$
Let

$z$ be the total profit.

Then,

$z=500x+300y$

$\therefore$ the given problem reduces to maximise the objectives functions

$z=500x+300y$ subject to the constraints

$x\ge 0$ ....(i)

$y\ge 0$ .....(ii)

$x+y\le 200$ .......(iii)

$3x+2y\le 450$ .....(iv)

$x\ge 0$ and

$y\ge 0$ represent the closed half-planes on the next of y-axis and above x-axis respectively.
The line corresponding to

$(iii)$ is

$x+y=200$

$(200,0)$ and

$(0,200)$ lie on

$(iii)$ . The origin lies in the half-plane of

$(iii)$ if

$0+0\le 200$ . Which is true.

$\therefore$ The closed half-plane on the next of y-axis and above x-axis respectively.

$\therefore$ the closed half-plane containing the origin in graph of

$(iii)$
The line corresponding to

$(iv)$ is

$3x+2y-450....(vi)$ .

$(150,0)$ and

$(0,225)$ lie on

$(vi)$ .
The origin does not lie on

$(iv)$ and it lies in the half-plane of

$(iv)$ if

$3(0)+2(0)\le 450$ , which is true. So, the closed half-plane containing the origin in the graph of

$(iv)$ .
The shaded bounded region in the feasible region of the given L.P.P we use cover point method to maximise

$z$ . The required of the feasible region are

$O(0,0),A(0,200),B(75,125)$ and

$C(150,0)$
At

$O(0,0),z=500(0)+300(0)=0$
At

$A(0,200),z=500(0)+300(200)=60000$
At

$B(75,125),z=500(75)+300(125)=37500+37500=75000$
At

$C(150,0),z=500(150)+300(0)=75000$

$\therefore$ Maximize value of

$z=75,000$ when either

$x=75,y=125$ or

$x=150,y=0$

$\therefore$ profit of airline is maximum (Rs.

$75000$ ) where either these are

$75$ first class ticket holders and

$125$ economy class ticket holders or there are only

$150$ first class ticket holders.

Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = x + 2y$
subject to $x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200; x, y \geq 0$ .

Objective function :

$Z=x+2y$

We have to minimize and maximize Z on constraints

$x+2y\geq 100$

$2x-y\leq 0$

$2x+y\leq 200$

$x\geq 0, y\geq 0$

After plotting all the constraints on coordinate plane, we get the feasible region as shown in the image.

Corner points	Value of $Z=x+2y$
(0,200)	400 (Maximum)
(0,50)	100 (Minimum)
(20,40)	100 (Minimum)
(50,100)	250

Hence, maximum value of Z will be 400 and minimum value of Z will be 100.

Since, Minimum of Z will occurs for all the points of line

$x+2y=100$ lying in the feasible region.

Hence, there are more than 2 values where minimum of Z will occur.

Solve the following Linear Programming Problems graphically:
Minimise $Z = x + 2y$
subject to $2x + y \geq 3, x + 2y \geq 6, x, y \geq 0$ .

Given objective function is

$Z=x+2y$

We have to minimize Z on constraints

$2x+y\geq 3$

$x+2y\geq 6$

$x\geq 0,y\geq 0$

After plotting the inequalities we got the feasible region as shown in the image

Now there are two corner points

$(0,3)\ and\ (6,0)$ lying on same line

$x+2y=6$

Value at corner points are :

Corner Points	Value of $Z=x+2y$
(0,3)	6 (minimum)
(6,0)	6 (minimum)

Since, feasible region is unbounded. So, 6 may or may not be minimum value.

Now to check if 6 is minimum or not, we have to draw

$Z<6\Rightarrow x+2y< 6$

Since this region doesn't have any common region with feasible region.

So, 6 is the minimum value of Z.

A factory manufactures two types of screws, $A$ and $B$ . Each type of screw requires the use of two machines, an automatic and a hand-operated. It takes $4$ minutes on the automatic and $6$ minutes on hand-operated machines to manufacture a package of screws $A$ , while it takes $6$ minutes on automatic and $3$ minutes on the hand-operated machines to manufacture a package of screws $B$ . Each machine is available for at the most $4$ hours on any day. The manufacturer can sell a package of screws $A$ at a profit of $Rs. 7$ and screws $B$ at a profit of $Rs. 10$ . Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

Let number of package of screws A to be made is

$X$ and number of package of screws B to be made is

$Y$

Since, screws A package requires

$4$ minutes and screws B package requires

$6$ minutes of automatic machine work time. Also, there is maximum

$4$ hours (240 minutes) of automatic machine time available.

$\therefore 4X+6Y\leq 240$

$\Rightarrow 2X+3Y\leq 120\ \ \ ...(1)$

Since, screws A package requires

$6$ minutes and screws B package requires

$3$ minutes of hand operated machine work time. Also, there is maximum

$4$ hours (240 minutes) of hand operated machine time available.

$\therefore 6X+3Y\leq 240$

$\Rightarrow 2X+Y\leq 80\ \ \ ...(2)$

Since, count of package can't be negative.

$\therefore X\geq 0,Y\geq 0\ \ \ ...(3)$

We have to maximize profit of the factory.

Here, profit on screws A package is

$7\ Rs$ and on screws B package is

$10\ Rs$

So, objective function is

$Z=7X+10Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=7X+10Y$
A $(0,40)$	$400$
B $(30,20)$	$410$ (maximum)
C $(40,0)$	$280$

Hence, factory owner should produce 30 packages of screws A and 20 packages of screws B in a day to maximise his profit. Also, maximum profit will be

$410\ Rs$

Reshma wishes to mix two types of food $P$ and $Q$ in such a way that the vitamin contents of the mixture contain at least $8\ units$ of vitamin $A$ and $11\ units$ of vitamin $B$ . Food $P$ costs $Rs.\ 60/kg$ and Food $Q$ costs $Rs.\ 80/kg$ . Food $P$ contains $3\ units/kg$ of Vitamin $A$ and $5\ units / kg$ of Vitamin $B$ while food $Q$ contains $4\ units/kg$ of Vitamin $A$ and $2\ units/kg$ of vitamin $B$ . Determine the minimum cost of the mixture.

Let's assume that mixture contains

$X\ kg$ of food P type and

$Y\ kg$ of food Q type.

Cost of 1 kg of food P

$=60\ Rs$

Cost of 1 kg of food Q

$=80\ Rs$

So, Total cost (C)

$=60X+80Y\ Rs\ \ \ \ ...(1)$

Now, food P contains

$3\ units/kg$ of vitamin A and food Q contains

$4\ units/kg$ of vitamin A.

So, Total Vitamin A contains

$=3X+4Y$

Since minimum requirement of Vitamin A is 8 units.

$\therefore 3X+4Y\geq 8\ \ \ \ ...(2)$

Also, food P contains

$5\ units/kg$ of vitamin B and food Q contains

$2\ units/kg$ of vitamin B.

So, Total Vitamin B contains

$=5X+2Y$

Since the minimum requirement of Vitamin B is 11 units.

$\therefore 5X+2Y\geq 11\ \ \ \ ...(3)$

Since the amount of food can never be negative.

So,

$X\geq 0, Y\geq 0\ \ \ \ ...(4)$

We have to minimize the cost of mixture given in equation (1) subject to the constraints given in (2),(3) and (4).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $Z=60X+80Y$
$\left (0,\dfrac{11}{2} \right )$	$440$ (Maximum)
$\left (2,\dfrac{1}{2} \right )$	$160$ (Minimum)
$\left (\dfrac{8}{3},0 \right )$	$160$ (Minimum)

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, plot the region

$Z< 160$ to check if there exist some points in the feasible region where value can be less than 160.

$\Rightarrow 60X+80Y<160$

$\Rightarrow 3X+4Y<8$ .

Since there is no common points between the feasible region and the region which contains

$Z<160$ . So 160 is the minimum cost.

Show that the maximum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = 5x + 10y$ subject to $x + 2y \leq 120, x + y \geq 60, x - 2y \geq 60, x - 2y \geq 0, x, y\geq 0$ .

Objective function :

$Z=5x+10y$

We have to minimize and maximize Z on constraints

$x+2y\leq 120$

$x+y\geq 60$

$x-2y\geq 60$

$x-2y\geq 0$

$x\geq 0, y\geq 0$

After plotting all the constraints on coordinate plane, we get the feasible region as shown in the image.

Corner points	Value of $Z=5x+10y$
(60,0)	300 (Minimum)
(90,15)	600 (Maximum)
(120,0)	600 (Maximum)

Here maximum of

$Z = 600$

Plotting the line

$Z=600 \Rightarrow 5x+10y=600$

Since this line coincide with line

$x+2y=120$ , Hence all the points lying on line

$x+2y=120$ in feasible region will give maximum value.

So, there are more than 2 values where maximum of Z will occur.

Show that the maximum of $Z$ occurs at more than two points.
Maximise $Z = -x + 2y$ , subject to the constraints:
$x\geq 3, x + y \geq 5, x + 2y \geq 6, y \geq 0$ .

Objective function :

$Z=-x+2y$

We have to minimize and maximize Z on constraints

$x\geq 3$

$x+y\geq 5$

$x+2y\geq 6$

$y\geq 0$

After plotting all the constraints on coordinate plane, we get the feasible region as shown in the image.

Corner points	Value of $Z=-x+2y$
(3,2)	1 (Maximum)
(4,1)	-2
(6,0)	-6 (Minimum)

Now, since region is unbounded. Hence we need to confirm that minimum and maximum value obtained through corner points is true or not.

Now, plot the region

$Z\leq -6\Rightarrow -x+2y\leq -6$ . Since, this region has common region with feasible region. Hence, there are many points where value of Z will be less than -6. So, minimum of Z not exists.

(Example :- check the value of Z at

$(8,0.5)$ )

Similarly, plot the region

$Z\geq 1\Rightarrow -x+2y\geq 1$ . Since, this region has common region with feasible region. Hence there are many points where value of Z will be greater than 1. So, maximum of Z doesn't exists.

(Example :- check the value of Z at

$(4,4)$ )

A manufacturing company makes two types of teaching aids A and B of Mathematics for Class X. Each type of A requires $9$ labour hours for fabricating and $1$ labour hour for finishing. Each type of B requires $12$ labour hours for fabricating and $3$ labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are $180$ and $30$ respectively. The company makes a profit of Rs. $80$ on each piece of type A and Rs. $120$ on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.

Let the pieces of type

$A$ manufactured per week be

$x$ .

Let the pieces of type

$B$ manufactured per week be

$y$ .

$\therefore$ maximum profit

$z=80x+120y$

Fabricating hours for

$A$ is

$9$ and finishing hours is

$1$

Fabricating hours for

$B$ is

$12$ and finishing hours is

$3$

Maximum number of fabricating hours is

$180$ .

$\therefore 9x+12y\leq 180$

$\Rightarrow 3x+4y\leq 60$

Maximum number of finishing hours is

$30$ .

$\therefore x+3y\leq 30$

Thus we can formulate the Linear Programming Problem as follows:

$max\:z=80x+120y$ subjected to the constraints,

$3x+4y\leq 60, x+3y\leq 30,x\geq 0,y\geq 0$

Solving the constraints we get

$x=12,y=6$

Now the area of feasible region is as shown in the figure.

We can see that the points bounded by the feasible region are

$A(0,10),B(12,6),C(20,0)$

Now let us calculate the maximum profit.

$A(0,10), z=80(0)+120(10)=1200$

$B(12,6), z=80(12)+120(6)=1680$

$C(20,0), z=80(20)+120(0)=1600$

Hence the maximum profit is at

$(12,6)$

Therefore

$12$ pieces of type

$A$ and

$6$ pieces of type

$B$ should be manufactured per week to get a maximum profit of

$Rs.1680$

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type $A$ require $5$ minutes each for cutting and $10$ minutes each for assembling. Souvenirs of type $B$ require $8$ minutes each for cutting and $8$ minutes each for assembling. There are $3$ hours $20$ minutes available for cutting and $4$ hours for assembling. The profit is $Rs. 5$ each for type $A$ and $Rs. 6$ each for type $B$ souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

Let number of souvenirs A to be made is

$X$ and number of souvenirs B to be made is

$Y$

Since, souvenirs A requires

$5$ minutes and souvenirs B requires

$8$ minutes for cutting. Also, there is maximum

$3$ hours

$20$ minutes (200 minutes) for cutting.

$\therefore 5X+8Y\leq 200\ \ \ ...(1)$

Since, souvenirs A requires

$10$ minutes and souvenirs B requires

$8$ minutes for assembling. Also, there is maximum

$4$ hours (240 minutes) for assembling.

$\therefore 10X+8Y\leq 240$

$\Rightarrow 5X+4Y\leq 120\ \ \ ...(2)$

Since, count of objects can't be negative.

$\therefore X\geq 0,Y\geq 0\ \ \ ...(3)$

We have to maximize profit of the company.

Here, profit on souvenirs A is

$5\ Rs$ and on souvenirs B is

$6\ Rs$

So, objective function is

$Z=5X+6Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=5X+6Y$
A $(0,25)$	$150$
B $(8,20)$	$160$ (maximum)
C $(24,0)$	$120$

Hence, company should produce 8 souvenirs A and 20 souvenirs B in a day to maximize his profit. Also, maximum profit will be

$160\ Rs$

Determine graphically the minimum value of the objective function.
$Z = -50x + 20y$
Subject to constraints
$2x - y \geq -5$
$3x + y \geq 3$
$2x - 3y \leq 12$
$x \geq 0, y \geq 0$ .

Given objective function is

$Z=-50x+20y$

We have to minimize Z on given constraints

$2x-y\geq -5$

$3x+y\geq 3$

$2x-3y\leq 12$

$x\geq 0,y\geq 0$

After plotting all the constraints we get the common region (Feasible region) as shown in the image.

There are four corner points

$(0,5),(0,3),(1,0)\ and\ (6,0)$

Now, at corner points value of Z are as follows :

Corner Points	Value of $Z=-50x+20y$
(0,5)	100
(0,3)	60
(1,0)	-50
(6,0)	-300 (minimum)

Since common region is unbounded. So, value of Z may be minimum at

$(6,0)$ and minimum value may be -300.

Now to check if this minimum is correct or not, we have to draw region

$-50x+20y\leq -300$

Since, there are some common region with feasible region(See image). So, -300 will not be minimum value of Z.

Hence, Z has no minimum value.

Use the graphical method to solve each of the following LP problems.
A wheat and barley farmer has $168$ hectare of ploughed land, and a capital of $E2000$ . It costs $E14$ to sow hectare wheat ad $E10$ to sow one hectare of barley.
Suppose that his profit is $E80$ per hectare of wheat and $E55$ per hectare of barley. Find the optimal number of hectares of wheat and barley that must be ploughed in order to maximise profit? What is the maximum profit?

assume that

$x$ is wheat and

$y$ is barley hectaters to maximum the profit

by question given that to convert in mathematically form

$x\times 14+10\times y \le 2000$

profit

$Z=80\times x+55y$

and

$x+y \le 168$

$x\le 0$

$y\le 0$

$7x+5y \le 1000$

$x+y \le168$

$x\le 0$

$y\le 0$

for

$7x+5y\le 1000$

x	5	135
y	193	11

for equation

$x+y\le 168$

x	0	168
y	168	0

by graphs

than pain A,B,C

point	lecible solution
A(0,200)	Z=E11,000
B(0,168)	Z=E92,40
C(80,88)	Z=11,240

Finally wheat hector is

$80$ and barley hector

$88$ than maximum profit is

$=11,240$

profit

$=E11240$

1063407_847777_ans_b827d18aa376447da20273d787d7d2ac.png

Solve the following minimal assignment problem and hence find the minimum value:
I II III IV
A 2 10 9 7
B 13 2 12 2
C 3 4 6 1
D 4 15 4 9

Table 1.

	I	II	III	IV
A	2	10	9	7
B	13	2	12	2
C	3	4	6	1
D	4	15	0	5

Table 2.

	I	II	III	IV
A	0	8	7	5
B	11	0	10	0
C	2	3	5	0
D	0	11	0	5

Table 3.

	I	II	III	IV
A	0	8	7	5
B	11	0	10	0
C	2	3	5	0
D	0	11	0	5

Table 1

Subtracting minimum element of each row from the row elements,

Table 2

Subtracting minimum element of each column we have,

Table 3

Now going anti clockwise and noting the zero positions, the cells picked are,

$A-I \quad B \quad II \quad\quad C-IV \quad D\quad III$

$\therefore$ minimum value=sum of values in these cells

$=2+2+1+4=9$

Answer

$=9$

An aeroplane can carry a maximum of $200$ passengers. A profit of $Rs. 1000$ is made on each executive class ticket and a profit of $Rs. 600$ is made on each economy class ticket. The airline reserves at least $20$ seats for executive class. However, at least $4$ times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

Let's assume that executive class ticket sold be X and economy class ticket sold be Y.

Since, aeroplane can carry maximum 200 passengers.

$\therefore X+Y\leq 200\ \ \ ...(1)$

Since, atleast 20 tickets is reserves for executive class.

$\therefore X\geq 20\ \ \ ...(2)$

Since the number of tickets for economy class should be at least 4 times the executive class.

$\therefore Y\geq 4X\Rightarrow Y-4X\geq 0\ \ \ ...(3)$

Also, the number of tickets can't be negative.

So,

$X\geq 0\ and\ Y\geq 0\ \ \ ...(4)$

Profit on an executive class ticket is

$1000\ Rs$ and profit on an economy class ticket is

$600\ Rs$

So, Total profit

$(Z)=1000X+600Y$

We have to maximize the total profit. After plotting all the constraints given by equation (1), (2), (3) and (4), we get the feasible region as shown in the image.

Corner points	Value of $Z=1000X+600Y$
A (20, 180)	128000
B (40, 160)	136000 (Maximum)
C (20,80)	68000

Hence, maximum profit will be

$136000\ Rs$ , when number of executive class ticket sold be

$40$ and number of economy class ticket sold be

$160$

A merchant plans to sell two types of personal computers - a desktop model and a portable model that will cost $Rs. 25000$ and $Rs. 40000$ respectively. He estimates that the total monthly demand of computers will not exceed $250$ units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than $Rs. 70$ lakhs and if his profit on the desktop model is $Rs. 4500$ and on portable model is $Rs. 5000$ .

Let number of desktop model be

$x$ and number of portable model be

$y$

According to question,

Since, monthly demand doesn't exceed 250 units.

$\therefore x+y\leq 250\ \ \ ...(1)$

Since, maximum invest is 70 lakhs.

$\therefore 25000x +40000y\leq 7000000\Rightarrow 5x+8y\leq 1400\ \ \ ...(2)$

Also, quantity can't be negative.

$\therefore x\geq 0, y\geq 0\ \ \ ...(3)$

We have to maximize profit

$Z$

where

$Z= 4500x +5000y$

After plotting all the constraints given by equation (1),(2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=4500x+5000y$
A(0,175)	875000
B(200,50)	1150000 (Maximum)
C(250,0)	112500

Hence, maximum profit that can be earned by the merchant will be

$1150000\ Rs$

Construct the graphs of the following functions.
$\displaystyle y \, = \, (x \, + \, 3) \, / \, (x \, - \, 1).$

according to question if

$x=1$ then the function is undefined

therefore the domain is

$R-{1}$

when x tends to infinity the value becomes 1

Hence, we get the above graph

Construct the graphs of the following functions.
$\displaystyle x \, = \, \sqrt{1 \, - \, | y |}.$
The maximum possible value of x is:

1958173_891123_ans_e0a493e05af746139d6a2986df346d34.jpg

A toy company manufactures two types of dolls, $A$ and $B$ . Market tests and available resources have indicated that the combined production level should not exceed $1200$ dolls per week and the demand for dolls of type $B$ is at most half of that for dolls of type $A$ . Further, the production level of dolls of type $A$ can exceed three times the production of dolls of other types by at most $600$ units. If the company makes a profit of $Rs. 12$ and $Rs. 16$ per doll respectively on dolls $A$ and $B$ , how many of each should be produced weekly in order to maximize the profit?

Let's assume that the number of dolls of type A is

$X$

and number of dolls of type B be

$Y$

Since combined production level should not exceed 1200 dolls

$\therefore X+Y\leq 1200\ \ \ ...(1)$

Since production levels of dolls of type A exceeds 3 times the production of type B by at most 600 units

$\therefore X-3Y\leq 600\ \ \ ...(2)$

Also, the demands of dolls of type B is at most half of that for dolls of type A

$\therefore Y\leq \dfrac{X}{2}$

$\Rightarrow 2Y-X\leq 0\ \ \ ...(3)$

Since the count of an object can't be negative.

So,

$X\geq 0,Y\geq 0\ \ \ ...(4)$

Now, profit on type A dolls

$=12\ Rs$

and profit on type B dolls

$=16\ Rs$

So, total profit

$Z=12X+16Y$

We have to maximize the total profit (Z) of the manufacturers.

After plotting all the constraints given by equation (1), (2), (3) and (4) we get the feasible region as shown in the image.

Corner points	Value of $Z=12X+16Y$
A (800,400)	16000 (maximum)
B (1050,150)	15000
C (600,0)	7200
O (0,0)	0

So, in order to maximize the profit, the company should produce

$800$ type A dolls and

$400$ type B dolls

A diet is to contain at least $80$ units of vitamin A and $100$ units of minerals. Two foods $F_{1}$ and $F_{2}$ are available. Food $F_{1}$ costs $Rs. 4$ per unit food and $F_{2}$ costs $Rs. 6$ per unit. One unit of food $F_{1}$ contains $3$ units of vitamin A and $4$ units of minerals. One unit of food $F_{2}$ contains $6$ units of vitamin A and $3$ units of minerals. Formulate this as a linear programming problem. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.

Let's assume that diet contains

$X\ units$ of food

$F_1$ and

$Y\ units$ of food

$F_2$ .

Cost of 1 unit of food

$F_1\ =4\ Rs$

Cost of 1 unit of food

$F_2\ =6\ Rs$

So, Total cost (C)

$=4X+6Y\ Rs\ \ \ \ ...(1)$

Now, food

$F_1$ contains

$3\ units$ of vitamin A and food

$F_2$ contains

$6\ units$ of vitamin A.

So, Total Vitamin A contains

$=3X+6Y$

Since, minimum requirement of Vitamin A is 80 units.

$\therefore 3X+6Y\geq 80\ \ \ \ ...(2)$

Also, food

$F_1$ contains

$4\ units$ of minerals and food

$F_2$ contains

$3\ units$ of minerals.

So, Total minerals contains

$=4X+3Y$

Since, minimum requirement of minerals is 100 units.

$\therefore 4X+3Y\geq 100\ \ \ \ ...(3)$

Since, amount of food can never be negative.

So,

$X\geq 0, Y\geq 0\ \ \ \ ...(4)$

We have to minimise the cost of diet given in equation (1) subject to the constraints given in (2), (3) and (4).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $C=4X+6Y$
A $(0,\dfrac{100}{3})$	$200$
B $(24,\dfrac{4}{3})$	$104$ (Minimum)
C $(\dfrac{80}{3},0)$	$106\dfrac{2}{3}$

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, plot the region

$Z< 104$ to check if there exist some points in feasible region where value can be less than 160.

$\Rightarrow 4X+6Y<104$

Since there is no common points between the feasible region and the region which contains

$Z<104$ . So 104 is the minimum cost.

A farmer mixes two brands $P$ and $Q$ of cattle feed. Brand $P$ , costing $Rs. 250$ per bag, contains $3$ units of nutritional element $A, 2.5$ units of element $B$ and $2$ units of element $C$ . Brand $Q$ costing $Rs. 200$ per bag contains $1.5$ units of nutritional element $A, 11.25$ units of element $B$ , and $3$ units of element $C$ . The minimum requirements of nutrients $A, B$ and $C$ are $18\ units, 45\ units$ and $24\ units$ respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Let's assume that mixture contains

$X\ bags$ of brands P and

$Y\ bags$ of brands Q.

Cost of 1 bag of brands P

$=250\ Rs$

Cost of 1 bag of brands Q

$=200\ Rs$

So, Total cost (C)

$=250X+200Y\ Rs\ \ \ \ ...(1)$

Now, brands P contains

$3\ units$ and brands Q contains

$1.5\ units$ of nutritional element A.

So, Total nutritional element A

$=3X+1.5Y\ units$

Since, minimum requirement of nutritional element A is 18 units.

$\therefore 3X+1.5Y\geq 18$

$\Rightarrow 2X+Y\geq 12\ \ \ \ ...(2)$ (After dividing by 1.5 both side)

Again, brands P contains

$2.5\ units$ and brands Q contains

$11.25\ units$ of nutritional element B.

So, Total nutritional element B

$=2.5X+11.25Y\ units$

Since, minimum requirement of nutritional element B is 45 units.

$\therefore 2.5X+11.25Y\geq 45$

$\Rightarrow 2X+9Y\geq 36\ \ \ \ ...(3)$ (After dividing by 1.25 both side)

Also, brands P contains

$2\ units$ and brands Q contains

$3\ units$ of nutritional element C.

So, Total nutritional element C

$=2X+3Y\ units$

Since, minimum requirement of nutritional element C is 24 units.

$\therefore 2X+3Y\geq 24\ \ \ \ ...(4)$

Since, amount of brands can never be negative.

So,

$X\geq 0, Y\geq 0\ \ \ \ ...(5)$

We have to minimise the cost of mixture given in equation (1) subject to the constraints given in (2),(3), (4) and (5).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $Z=250X+200Y$
A (0,12)	2400
B (3,6)	1950 (minimum)
C (9,2)	2650
D (18,0)	4500

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, plot the region

$Z< 1950$ to check if there exist some points in feasible region where value can be less than 1950.

$\Rightarrow 250X+200Y<1950$

$\Rightarrow 5X+4Y<39$ .

Since there is no common points between the feasible region and the region which contains

$Z<1950$ (See the image). So 1950 is the minimum cost.

An company manufactures two electrical products: air conditioners and large fans. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes $3$ hours of wiring and $2$ hours of drilling. Each of fan must go through $2$ hours of wiring and $1$ hour of drilling. During the next production period, $240$ hours of wiring time are available and up to $140$ hours of drilling time may be used.
Each air conditioner sold yields a profit of $E25$ . Each fan assembled may be sold for a profit of $E15$ . Formulate and solve this linear programming mix situation to find the best combination of air conditioners and fans that yields the highest profit.

Assume that

$x$ is number of airconditioners

$y$ is number of fans units.

according to question.

wiring

$\Rightarrow 3x+1y=240$ .......(1)

Drilling

$\Rightarrow 2x+y\le 140$ .......(2)

and profit

$z=25x+15y$

minimum air conditioners

$x\ge 20$

maximum fans

$0\le y\le 80$

by equation (1) and (2)

$3x+y \le 240$

for the graph

$x$ will be

$x$ and

$y$ will be

$y$ (Refer image)

$(20,2)$

$(70,0)$

$(20,80)$

$(80,80)$

there four vertices define the feasible region. check the value of the profit function at these four points to find the maximum value at

$(20, 0)$

$z=25\times 20+0$

$=500$

$(70,0)$

$Z=1750$

$(20,80)z=1700$

$(20,80)z=1950$

maximum profit

$z=1950$

1448245_847788_ans_a3bea894e7a14c62a88f18ec60a7b8b1.png

Solve graphically:
$x+2y\le 10;x+y\ge 1;x-y\le 0;\le 0$

First we solve

$X+2Y\le10$

Lets first draw graph of

$X+2Y=10$

Putting

$X+10$

$0+2Y=10$

$2Y=10$

$Y=5$

And then,Putting

$Y=0$

$X+2(0)=10$

$X+0=10$

$X=10$

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most $24$ . It takes one hour to make a bracelet and a half an hour to make a necklace. The maximum number of hours available per day is $16$ . If the profit on a necklace is $Rs\100$ and that on a bracelet is $Rs\ 300$ . Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.

Let $x$ be the number of necklaces manufactured and $y$ be the number of bracelets manufactured. The total number of necklaces and bracelets it can handle is at most 24, so, according to the question,

$x + y \le 24$

The $x$ items take $x$ hours to manufacture and $y$ items take $\frac{y}{2}$ hours to manufacture and the maximum time available is for 16 hours. So,

$x + \dfrac{y}{2} \le 16$

The profit on one necklace is given as ${\rm{Rs}}\;100$ and the profit on one bracelet is given as ${\rm{Rs}}\;300$ .

Let the profit be $z$ . To maximize the profit,

$z = 100x + 300y$

Therefore, the required constraints is,

$z = 100x + 300y$

Subject to the constraints,

$x + y \le 24$

$x + \dfrac{y}{2} \le 16$

An advertising company wishes to plan its advertising strategy in three different media-television,radio and magazines.The purpose of advertising is to reach as large a number of potential customers as possible.Following data have been obtained from market survey:
Strategies Television Radio Magazine-I Magazine-II
Cost of an advertising company Rs. $30,000$ Rs. $20,000$ Rs. $15,000$ Rs. $10,000$
No.of potential customers reached per unit $2,00,000$ $6,00,000$ $1,50,000$ $1,00,000$
No.of female customers reached per unit $1,50,000$ $4,00,000$ $70,000$ $50,000$
The company wants to spend not more than Rs. $4,50,000$ on advertising.Following are the further requirements that must be met:
$\left(i\right)$ at least $1$ million exposures take place among female customers.
$\left(ii\right)$ advertising on magazines be limited to Rs. $1,50,000$
$\left(iii\right)$ at least $3$ advertising units be bought on magazine $I$ and $2$ units on magazine $II$
$\left(iv\right)$ the number of advertising units on television and radio should each be between $5$ and $10.$
Formulate an L.P model for the problem.

Formulation of Linear programming problem:
Let

${x}_{1},{x}_{2},{x}_{3}$ and

${x}_{4}$ denote the number of advertising units to be bought on television,radio,magazine

$I$ and magazine

$II$ respectively.
The objective is to maximize the total number of potential customers reached.
i.e., Maximize

$Z={10}^{5}\left(2{x}_{1}+6{x}_{2}+1.5{x}_{3}+{x}_{4}\right)$
Constraints are
on the advertising budget:

$30,000{x}_{1}+20,000{x}_{2}+15,000{x}_{3}+10,000{x}_{4}\le 4,50,000$
or

$30{x}_{1}+20{x}_{2}+15{x}_{3}+10{x}_{4}\le 450$
on the number of female customers reached by the advertising campaign:

$1,50,000{x}_{1}+4,00,000{x}_{2}+70,000{x}_{3}+50,000{x}_{4}\ge 10,00,000$
or

$15{x}_{1}+40{x}_{2}+7{x}_{3}+5{x}_{4}\ge100.$
on expenses on magazine advertising:

$15,000{x}_{3}+10,000{x}_{4}\le1,50,000$
or

$15{x}_{3}+10{x}_{4}\le 150$
on no. of units on magazines:

${x}_{3}\ge3,{x}_{4}\ge2$
on no. of units on television:

$5\le{x}_{1}\le10$ or

${x}_{1}\ge5,{x}_{1}\le10,$
on no. of units on radio:

$5\le{x}_{2}\le10$ or

${x}_{2}\ge5,{x}_{2}\le10,$
where

${x}_{1},{x}_{2},{x}_{3},{x}_{4},$ each

$\ge0.$

Solve the following system of inequalities graphically $x\ge 3, y\ge 2$

The set of inequalities are given as,

$x \ge 3$

$y \ge 2$

The equation $x = 3$ will give a straight line and since $0 \ge 3$ is not true, so the graph will be shaded away the origin.

In the same way, $y = 2$ will give a straight line and since $0 \ge 2$ is not true, so the graph will be shaded away the origin.

1004152_1064084_ans_06b665e9214844499f1960ab6c113d84.png

David wants to invest at most $Rs.\ 12,000$ in bonds $A$ and $B$ . According to the rule, he has to invest at least $Rs.\ 2,000$ in Bond $A$ and at least $Rs.\ 4,000$ in Bond $B$ . If the rates of interest on Bonds $A$ and $B$ respectively are $8\%$ and $10\%$ perannum, formulate the problem as L.P.P. and solve it graphically for maximum interest. Also determine the maximum interest received in a year.

Let David invested

$Rs\ x$ &

$Rs\ y$ in Bond

$A$ & Bond

$B$ resp

Rate of interest on Bond

$A=8\% /annum$

$B=10\% /annum$

Interest on Bond

$A=x 8/100$

$=Rs\ 2x/25$

$B=y 10/100$

$=Rs\ y/10$

Total Interest

$=Rs\ \dfrac{2x}{25}+\dfrac{y}{10}$

$z=\dfrac{2x}{25}+\dfrac{y}{10}$

Constants:

$x+y\le 12,000$

$x\ge 2000$

$y\ge 4000$

$A(2000, 4000); B(8000, 4000)$

$C(2000, 10000)$

$A$

$Z=\dfrac{2(2000)}{25}+\dfrac{4000}{10}=160+400=560$

$B$

$z=\dfrac{2(8000)}{25}+\dfrac{4000}{10}=640+400=1040$

$C$

$z=\dfrac{2(2000)}{25}+\dfrac{10,000}{10}=160+1000=1160$

$\rightarrow z=-1160$ (max value)

He must invest

$Rs\ 2000$ in Bond

$A, 10,000$ in Bond

$B$

1381648_1140525_ans_3454674faaa64a84a56101e2cd9da009.png

Find the number of integer solutions of ${x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}=20$ where ${x}_{1}\ge -5, {x}_{2}\ge 3, {x}_{3}\ge 0, {x}_{4}\ge 1$

$x_{1}+ x_{2}+ x_{3}+ x_{4} = 20$

$x_{1} \ge -5 \Rightarrow x_{1}+5 \ge 0$

$x_{2} > 3 \Rightarrow x_{2}-3 \ge 0$

$x_{3} \ge 0$

$x_{4} \ge 1 \Rightarrow x_{4}- 1 \ge 0$

Let

$x_{1}+5 = k, x_{2}-3 = k_{2} , x_{3}= k_{3} , x_{4}-1= k_{4}$

$k_{1}+ k_{2}+ k_{3}+ k_{4} = x_{1}+5 + x_{2}-3 + x_{3}+ x_{4}-1$

$= x_{1}+ x_{2}+ x_{3}+ x_{4}+1$

$k_{1}+ k_{2}+ k_{3}+ k_{4} = 20+1=21$

No. of non negative integral solutions of

$eq (2)$ is solution

i.e.,

$^{n+r-1}C_{r-1} [n= 21, r=4]$

$= ^{21+4-1}C_{4-1} = ^{24}C_{3}= 2024$

The following table gives production yields per hutone whest of 100 farms of a village
Production Yield 50-55 55-60 60-65 65-70 70-75 75-80
Number of Farms 2 8 12 24 38 46

Change the distribution to a more then type distribution and draw its vgive.

2041556_1056021_ans_f669ec8a2f9d4790ab2ec5f04a91b88b.jpg

To solve the following L.P.P.
Minimize $z=30x+20y$
Subject to the constraint, $x+y\leq 8$ , $x+2y \geq 4$ , $6x+7y\geq 12$ , $x\geq 0; y\geq 0$ .

$\begin{array}{l} Minimize=z=30x+20y \\ x+y\le 8,\, \, x+2y\ge 4,\, \, 6x+7y\ge 12,\, \, x+y\ge 0 \\ x+y=8 \\ put\, \, x=0,1,2 \\ \Rightarrow y=8,7,6\, \, respectively \\ x+2y=1 \\ put\, \, x=0,2,4 \\ y=2,1,0\, \, respectively \\ and \\ \Rightarrow 6x+7y=12 \\ \Rightarrow put\, \, x=0,\, \, y=\frac { { 12 } }{ 7 } \\ \Rightarrow put\, x=1,\, \, =\frac { { 12 } }{ 7 } \\ \Rightarrow put\, \, x=2,\, y=0 \end{array}$

hence minimum value of number at (0,2)

1216637_1137382_ans_0dab12b88a984417b16131dbdbbf7fb9.PNG

Solve the following L.P.P. graphically .
Subject to constraints
$2x+3y \ge 12$
$-x+y\le3$
$x\le 4$
$y\ge 3$

Corner points are

$(4,3),(4,7),(\dfrac{3}{2},3)(\dfrac{3}{5},\dfrac{18}{5}$

$(4,3),Z=3{x}+5{y}=12+15=27$

$(4,7),z=3{x}+5{y}=12+35=47$

$(\dfrac{3}{2},3),Z=3{x}+5{y}=\dfrac{9}{2}+15=\dfrac{39}{2}=19.5$

$(\dfrac{3}{5},\dfrac{18}{5}),Z=3{x}+5{y}=\dfrac{9}{5}+18=19.8$

Z is minimum at

$\bigg(\dfrac{3}{2},3\bigg)$

1173585_1424215_ans_7608bd316ba542d8bb8c8bfabfb8ea03.jpg

Minimize and maximize z=3x+9y subject to the constraints
$x+3y\le 60$
$10\le x+y$
$x\le y$
$0\le x,y$
by the graphical method. Minimum and maximum values are p and q then p+q =

1958182_1213307_ans_07362c8f74e2479ba1de94630675218c.jpg

A manufacturer makes two types of toys $A$ and $B$ . Three machine are need for this purpose and the time (In minutes) required for each toy on the machines is given below
Types of toys $I$ $II$ $III$
$A$ $20$ $10$ $10$
$B$ $10$ $20$ $30$
The machine $I,II$ and $III$ are available for a machine of $3$ hours, $2$ hours and $2$ hours $30$ minutes respectively. The profit on each on each toy of type $A$ is $Rs. 50$ and that of type $B$ is $Rs.\ 60$
Formulate the above problem as a LPP and solve it graphically to maximize profit.

1121371_1174459_ans_988d5b26872d4ef7bd48b9fd5648b51f.jpg

Solve each of the following linear programming problem by graphical method.
Maximize $Z=50x+30y$
Subject to $2x+y\le 18$
$3x+2y\le 34$
$x,y\ge 0$

1619464_1664172_ans_0bf6a24cd1f74a3bab255cdea4df261e.jpeg

A manufactures can produce two products, $A$ and $B$ , during a given time period. Each of these products requires four different manufacturing operation: grinding, turning, assembling and testing. The manufacturing requirements in hours per unit of products $A$ and $B$ are given below.
$A$ $B$
Grinding $1$ $2$
Turning $3$ $1$
Assembling $6$ $3$
Testing $5$ $4$
The available capacities of these operations in hours for the given time period are: grinding $30$ ; turning $60$ assembling $200$ ; testing $200$ . The contribution to profit is Rs. $2$ for each unit of $A$ and Rs. $3$ for each unit of $B$ . The firm can sell all that it produces at the prevailing market price. Determine the optimum amount of $A$ and $B$ to produce during the given time period. Formulate this as a LPP.

2039709_1664028_ans_a9e52a82d6e8454cb9826a929f0d51bc.png

Solve each of the following linear programming problem by graphical method.
Minimize $Z=5x+3y$
Subject to $2x+y\ge 10$
$x+3y\ge 15$
$x\le 10$
$y\le 8$
$x, y\ge 0$

2001916_1664223_ans_647d4006650a4e4da26c3a381fd68707.jpg

A company produces two types of goods, $A$ and $B$ that require gold and silver. Each unit of type $A$ requires $3$ gm of silver and $1$ gm of gold while that of type $B$ requires $1$ gm of silver and $2$ gm of gold. The company can produce $9$ gm of silver and $8$ gm of gold. If each unit of type $A$ brings a profit of Rs. $40$ and that of type $B$ Rs. $50$ , find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?

2039720_1664780_ans_08cf81dda07f4d18a1931d819ffee771.png

A furniture manufacturing company plans to make two products: chairs and tables. From its available resources which consists of $400$ square feet of teak wood and $450$ man hours. It is known that to make a chair requires $5$ square feet of wood and $10$ man-hours and yields a profit of Rs. $45$ , while each table uses $20$ square feet of word and $15$ man-hours and yields a profit of Rs. $80$ . How many items of each product should be produced by the company so that the profit is maximum?

2039721_1664795_ans_91ed6c8fafb14d16a5a5d2577cd991bf.png

Two tailors, $A$ and $B$ earn $Rs\ 15$ and $Rs\ 20$ per day respectively. A can stitch $6$ shirts and $4$ pants while $b$ can stitch $10$ shirts and $4$ pants per day. How many days shall each work if it is desired to produce (at least) $60$ shirts and $32$ pants at a minimum labour cost?

1606418_1664638_ans_6870d096187e404f994e22622f14ef06.jpg

A manufacturer makes two types $A$ and $B$ of tea-cups.Three machines are needed for the manufacturer and the time in minutes required for each cup on the machines is given below:
Machines
$I$ $II$ $III$
$A$ $12$ $18$ $6$
$B$ $6$ $0$ $9$
Each machine is available for a maximum of $6$ hours per day. If the profit on each cup $A$ is $75$ paise and that on each cup $B$ is $50$ paise, show that $15$ tea-cups of type $A$ and $30$ of type $B$ should be manufacturer in a day to get the maximum profit.

1573751_1664723_ans_255c43da2dd4419c8acc9a8d090a9c9a.png

A fruit grower can use two types of fertilizer in his garden, brand $P$ and $Q$ . The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given as at least $240$ kg of phosphoric acid, $270$ kg of potash and at most $310$ kg of chlorine.
$kg$ per bag
Brand $P$ Brand $Q$
Nitrogen $3$ $3.5$
Phosphoric acid $1$ $2$
Potash $3$ $1.5$
Chlorine $1.5$ $2$
If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

2039716_1664584_ans_2fba1fa406a3401cafd7a65143fcc767.jpg

A manufacturer of Furniture make two products : chairs and tables. Processing of these products is done on two machine $A$ and $B$ . A chair requires $2$ hr on machine $A$ and $6$ hrs on machine $B$ . A table requires $4$ hrs on machines $A$ and $2$ hrs on machine $B$ . There are $16$ hrs of time per day available on machine $A$ and $30$ hrs on machine $B$ . Profit gained by the manufacturer from a chair and a table is Rs. $3$ and Rs. $5$ respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his profit.

1961083_1664790_ans_e1847a51df324673bfe1e6e86918490e.jpg

A company produces two types of leather belts, say type $A$ and $B$ . Belt $A$ is a superior quality and belt $B$ is of a lower quality. Profits on each type of belt are Rs. $2$ and Rs. $1.50$ per belt, respectively. Each belt of type $B$ , the company could produce $1000$ belts per day. But the supply of leather is sufficient only for $800$ belts per day (both $A$ and $B$ combined). Belt $A$ requires a fancy buckle and only $400$ fancy buckles are available for this per day. For belt of type $B$ , only $700$ buckles are available per day. How should the company manufacture the two types of belts in order to have a maximum overall profit?

1610982_1664692_ans_69709e7580f34cecbdf989be13d90bd1.jpg

If a young man drives his scooter at a speed of $25$ km/hr, he to spend Rs. $2\ \text{per km}$ on petrol. If he drives the scooter at a speed of $40$ km/hour, it produces air pollution and increases his expenditure on petrol to Rs. $5\ \text{per km}$ . He has a maximum of Rs. $100$ to spend on petrol and travel a maximum distance in one hour time with less polution. Express this problem as an LPP and solve it graphically. What value do you find here?

2039717_1664606_ans_31cbdb3464704edf81fe333f5609b26e.png

A factory owner purchase two types of machines, $A$ and $B$ for his factory. The requirements and limitations for the machines are as follows:
Area occupied by the machine Labour force for each machine Daily output in units
Machine $A$ $1000.sq.\ m$ $12\ men$ $60$
Machine $B$ $1200\ sq.\ m$ $8\ men$ $40$
He has an area of $7600$ sq.m available and $72$ skilled men who can operate the machines. How many machines of each type should he buy to maximize the daily output?

2039719_1664765_ans_9b32b6651493480794f49ad774ff3903.png

A factory uses three different resources for the manufacture of two different products $20$ units of the resources $A, 12$ units of $B$ and $16$ units of $C$ being available. $1$ unit of the first produce requires $2,2$ and $4$ units of the respective resources and $1$ unit of the second product requires $4,2$ and $0$ units of respective resources. It is known that the first product gives a profit of $2$ monetary units per unit and the second $3$ . Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.

2005719_1664835_ans_87d80c18e22d45d8a1562eaa322b7b83.jpg

A man owns a field of area $1000$ sq.m. He wants to plant fruit trees in it. He has a sum of Rs. $1400$ to purchase young trees. He has the choice of two years. Type $A$ requires $10$ sq.m of ground per tree and costs Rs. $20$ per tree and type $B$ requires $20$ sq.m of ground per tree and costs Rs. $25$ per tree. When fully grown type $A$ produces an average of $20$ kg of fruit which can be sold at a profit of Rs. $2.00$ per kg and type $B$ produces an average $40$ kg of fruit which can be sold at a profit of Rs. $1.50$ per kg. How many of each type should be planted to achieve maximum profit when when the trees are fully grown? What is the maximum profit?

Total area of the field =

$1000 m^2$

Total sum of money the man has=

$Rs. 1400$

Let the number of type

$A$ and

$B$ trees he bought be

$x$ and

$y$ respectively.

Each type

$A$ tree requires

$10m^2$ of ground and costs

$Rs.20/$ tree.

Total area required by

$x$ type

$A$ trees =

$10x$

$m^2$

Total cost of

$x$ type

$A$ trees =

$Rs.20x$

Each type

$B$ tree requires

$20m^2$ of ground and costs

$Rs.25/$ tree.

Total area required by

$y$ type

$B$ trees =

$20y$

$m^2$

Total cost of

$y$ type

$B$ trees =

$Rs.25y$

Each fully grown type

$A$ tree produces

$20kg$ of fruit which can be sold at

$Rs.2/kg$ and each type

$B$ produces

$40kg$ of fruit which can be sold at

$Rs.1.50/kg$

Total income from all the trees=

$20 \times x \times 2 +40 \times y \times 1.50=40x+60y$

The required constraints are-

$10x+20y \leq 1000$

$. . . . . . . . . . . . . (i)$

$20x+25y \leq 1400$

$. . . . . . . . . . . . . (ii)$

$x \geq 0, y\geq 0$

$. . . . . . . . . . . . . (iii)$

Max

$Z$ =

$40x+60y$

Plotting the graphs of equations

$(i),(ii),(iii)$ we get the following corner points-

$A(20,40)$ ;

$B(0,50)$ ;

$C(70,0)$

Now

$Z$ is maximised at one of these corner points. Checking the value of

$Z$ for all these points we get-

$A$ ,

$Z=40 \times 20 + 60 \times 40= 800 +2400=3200$

$B$ ,

$Z= 40 \times 0 + 60 \times 50 = 3000$

$C$ ,

$Z = 40 \times 70 + 60 \times 0 = 2800$

Thus,

$Z$ is maximum at

$A(20,40)$

Hence,

$20$ type

$A$ trees and

$40$ type

$B$ trees should be bought to maximize the profit. And the maximum profit generated will be

$Rs.3200-Rs.1400 =Rs.2800$

A producer has $30$ and $17$ units of labour and capital respectively which he can use to produce two type of goods $X$ and $Y$ . To produce one unit of $X, 2$ units of labour and $3$ units of capital are required. Similarly $3$ units of labour and $1$ unit of capital is required to produce one unit of $Y$ . If $X$ and $Y$ are priced at Rs. $100$ and Rs. $120$ per unit respectively, how should be producer use his resources to maximize the total revenue? Solve the problem graphically.

Total units of labour = 30

Total units of capital = 17

Consider that

$x$ units of product

$X$ is made and

$y$ units of product

$Y$ is made.

One unit of

$X$ requires 2 units of labour and 3 units of capital.

Thus total units of labour required to make

$x$ units of

$X=$

$2x$ units

Total capital required to make

$x$ units of

$X=$

$3x$ units

One unit of

$Y$ requires 3 units of labour and 1 unit of capital.

Thus total units of labour required to make

$y$ units of

$Y=$

$3y$ units

Total capital required to make

$y$ units of

$Y=$

$y$ units

Price of

$X$ =

$Rs 100/$ unit

Price of

$Y$ =

$Rs.120/$ unit

Total revenue for selling both products

$=100x+120y$

The required constraints are-

$2x+3y \leq 30$

$. . . . . . . . . . . . . (i)$

$3x+y \leq 17$

$. . . . . . . . . . . . . (ii)$

$x \geq 0, y \geq 0$

$. . . . . . . . . . . . . (iii)$

Max

$Z=100x+120y$

Plotting the equations

$(i),(ii),(iii)$ we get the following corner points on the graph-

$A(3,8)$ ;

$B(0,10)$ ;

$C(\frac{17}{3},0)$

Now

$Z$ is maximized at any one of these three points.

$A$ ,

$Z=100 \times 3 + 120 \times 8 = 300 + 960 =1260$

$B$ ,

$Z=100 \times 0 + 120 \times 10 =1200$

$C$ ,

$Z=100 \times \frac{17}{3} + 120 \times 0 = 566.67$

Thus,

$Z$ is maximum at

$A(3,8)$

Thus the producer should produce

$3$ units of

$X$ and

$8$ units of

$Y$ to get the maximum revenue.

1620564_1665223_ans_794a4c2131084a0a887a7bfd55e324b1.PNG

A manufacture consider that men and women worker are equally efficient and so he pays them at the same rate. He has $30$ and $17$ units of workers (male and female) and capital respectively, which he uses to produces two types of goods $A$ and $B$ . To produce one unit of $A, 2$ workers and $3$ units of capital are required while $3$ workers and $1$ unit of capital is required to produce one unit of $B$ . If $A$ and $B$ are priced at Rs. $100$ and Rs. $120$ per unit respectively, how should he use his resource to maximize the total revenue? From the above as an LPP and solve graphically. Do you agree with this total view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?

1637315_1665533_ans_75772066a82d48419ee6a559219a6f47.jpg

A furniture dealer deals in only two items: tables and chairs. He has RS $30000$ to invest and a space to store at most $60$ pieces. A table costs him RS $1500$ and a chair RS $300.$ Formulate the data in the of inequatins and draw a graph representing the solution of these inequations.

1852742_1709039_ans_beacfff416f24e83bf2eac20c738e3ca.jpg

A manufacturer makes two products $A$ and $B$ . Product $A$ sells at Rs. $200$ each and takes $\dfrac {1}{2}$ hour to make. Product $B$ sells at Rs. $300$ each and takes $1$ hour to make. There is a permanent order for $14$ of product $A$ and $16$ of product $B$ . A working week consists of $40$ hours of production and weekly turnover must not be less than Rs. $10000$ . If the profit on each of product $A$ is Rs. $20$ and on product $B$ is Rs. $30$ , then how many of each should be produced so that the profits is maximum. Also, find the maximum profit.

2039722_1664900_ans_cbc1991f2f754a22a0d4980bddba6332.png

A firm manufactures two products $A$ and $B$ . Each product is processed on two machines $M_{1}$ and $M_{2}$ . Product A requires $4$ minutes of processing time on $M_{1}$ and $8$ min. on $M_{2}$ ; product $B$ requires $4$ minutes on $M_{1}$ and $4$ min. on $M_{2}$ . The machine $M_{1}$ is available for not more than $8$ hrs $20$ min. while machine $M_{2}$ is available for $10$ hrs during any working day. The products $A$ and $B$ are sold at a profit of Rs. $3$ and Rs. $4$ respectively.
Formulate the problem as a linear programming problem and find how many product of each type should be produced by the firm each day in order to get maximum profit.

2005717_1664809_ans_f674ced9877a47f5b73bae169d991972.jpg

A firm manufacturing two types of electric items, $A$ and $B$ can make a profit of Rs. $20$ per unit of $A$ and Rs. $30$ per unit of $B$ . Each unit of $A$ requires $3$ motors and $4$ transformers and each unit of $B$ requires $2$ motors and $4$ transformers. The total supply of these per month is restricted to $210$ motors and $300$ transformers. Type $B$ is an export model requiring a voltage stabilizer which has a supply restricted to $65$ units per month. Formulate the linear programming problem for maximum profit and solve ir graphically.

2005718_1664827_ans_4d5b94df635f4934bf8456b19751ae2c.jpg

A chemical company produces two compounds $A$ and $B$ . The following table gives the units of ingredients, $C$ and $D$ per $kg$ of compounds $A$ and $B$ as well as minimum requirement of $C$ and $D$ and costs per kg of $A$ and $B$ . Find the quantities of $A$ and $B$ which would give a supply of $C$ and $D$ at a minimum cost.
Compounds Minimum requirement
$A$ $B$
Ingredient $C$ $1$ $2$ $80$
Ingredient $D$ $3$ $1$ $75$
Cost (in $Rs$ ) $per\ kg$ $4$ $6$

2005720_1664878_ans_3640786561ca47b29db75c185d9f17ee.jpg

Two godowns, $A$ and $B$ grain storage capacity of $100$ quintals and $50$ quintals respectively. They supply to $3$ ration shops, $D, E$ and $F$ whose requirement are $60,50$ and $40$ quintals respectively. The cost transportation per quintal from the godowns to the shops are given in the following table:

Transportation cost per quintal (in Rs)
To $\downarrow$ \ From $\to$ $A$ $B$
$D$ $6.00$ $4.00$
$E$ $3.00$ $2.00$
$F$ $2.50$ $3.00$
How should the supplies be transported in order that the transportation cost is minimum?

1573875_1665579_ans_19b3f77e63224d308b883f343a6746b8.png

To receive grade $A$ in a course one must obtain an average of $90$ marks or more in five papers, each of $100$ marks. If Tanvy scored $89, 93, 95$ and $91$ marks in first four papers, find the minimum marks that she must score in the last paper to get grade $A$ in the course.

1982457_1709462_ans_f2cc1e14ff604068824ef3c91eb71840.jpg

Cutting alternatives	Width of Rolls(cm) $3$	Width of Rolls(cm) $4$	Width of Rolls(cm) $6$	Width of Rolls(cm) $10$	Waste(cm)
$1$	$4$	$3$	-	-	-
$2$	-	$3$	$2$	-	-
$3$	$1$	$1$	$1$	$1$	$1$
$4$	-	-	$2$	$1$	$2$
$5$	-	$4$	$1$	-	$2$
$6$	$3$	$2$	$1$	-	$1$

Items	Number of hours required on machines
	I	II	III
$M$	$1$	$2$	$1$
$N$	$2$	$1$	$1.25$

Types of Toys	Machines	Machines	Machines
	I	II	III
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

Transportation cost per quintal (in Rs)
From/ To	$A$	$B$
$D$	$6$	$4$
$E$	$3$	$2$
$F$	$2.50$	$3$

Kg per bag
	Brand $P$	Brand $Q$
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$

Ingredient	Brand Y Feed	Brand Z Feed	Minimum Monthly Requirement
A	$5$ oz	$10$ oz	$90$ oz
B	$4$ oz	$3$ oz	$48$ oz
C	$.5$ oz	$0$	$1.5$ oz
Cost/lb	$$0.2$	$$.03$

Branch Manager		Monthly Business (Rs. lakh)
	$A$	$B$	$C$	$D$
$P$	$11$	$11$	$9$	$9$
$Q$	$13$	$16$	$11$	$10$
$R$	$12$	$17$	$13$	$8$
$S$	$16$	$14$	$16$	$12$

Machines			Processing time in hours
	$A$	$B$	$C$	$D$	$E$
$M_{1}$	$9$	$11$	$15$	$10$	$11$
$M_{2}$	$12$	$9$	-	$10$	$9$
$M_{3}$	-	$11$	$14$	$11$	$7$
$M_{4}$	$14$	$8$	$12$	$7$	$8$

Machine	Time per unit (in minutes) of product- $1$	Time per unit (in minutes) of product- $2$	Time per unit (in minutes) of product- $3$	Machine capacity (minutes/day)
${M}_{1}$	$2$	$3$	$2$	$440$
${M}_{2}$	$4$	-	$3$	$470$
${M}_{2}$	$2$	$5$	-	$430$

Property	Property of raw material A	Property of raw material B	Property of raw material C
Specific gravity	$0.92$	$0.97$	$1.04$
Chromium	$7$ percent	$13$ percent	$16$ percent
Melting point	${440}^{0}$ Celsius	${490}^{0}$ Celsius	${480}^{0}$ Celsius

Food Type	Proteins (yield per unit)	Fats (yield per unit)	Carbohydrates (yield per unit)	Cost per unit (in Rupees)
$1$	$3$	$2$	$6$	$45$
$2$	$4$	$2$	$4$	$40$
$3$	$8$	$7$	$7$	$85$
$4$	$6$	$5$	$4$	$65$
Minimum Requirement	$800$	$200$	$700$

Gadget	Foundry	Machine-shop
$A$	$10$	$5$
$B$	$6$	$4$
Firm's capacity per week	$1000$	$600$

	Vitamin $A$	Vitamin $B$	Vitamin $C$	Vitamin $D$
Food $X$	$1$	$1$	$1$	$2$
Food $Y$	$2$	$1$	$3$	$1$

	Transportation cost per packet (in Rs)
To $\downarrow$ \ From $\to$	$A$	$B$
$P$	$5$	$4$
$Q$	$4$	$2$
$R$	$3$	$5$

		Cost of transportation (in Rs. per $1000$ bricks)
From To	A	B	C
$P$	$40$	$20$	$30$
$Q$	$20$	$60$	$40$

	Brand $P$	Brand $Q$
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$

Type of toys	Machine $I$	Machine $II$	Machine $III$
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

Food	Vitamin $A$	Vitamin $B$	Vitamin $C$
$X$	$1$ unit	$2$ units	$3$ units
$Y$	$2$ units	$2$ units	$1$ unit

Strategies	Television	Radio	Magazine-I	Magazine-II
Cost of an advertising company	Rs. $30,000$	Rs. $20,000$	Rs. $15,000$	Rs. $10,000$
No.of potential customers reached per unit	$2,00,000$	$6,00,000$	$1,50,000$	$1,00,000$
No.of female customers reached per unit	$1,50,000$	$4,00,000$	$70,000$	$50,000$

Production Yield	50-55	55-60	60-65	65-70	70-75	75-80
Number of Farms	2	8	12	24	38	46

Linear Programming - Class 12 Commerce Maths - Extra Questions

Solve the following equation2(2x+3)−10≤6(x−2)\displaystyle 2\left ( 2x+3 \right )-10\leq 6\left ( x-2 \right )

Check, whether the half plane 3x+6y≥03x+6y \geq 0 contains (1,1)(1, 1), if so shade the plane containing (1,1)(1, 1).

Shade the following half plane: 5x+10y−20≥05x+10y-20\geq 0.

Check, whether the half plane x+y+4>0x+y+4 > 0 contains the origin. Identify the plane and shade it.

The solution set of constraints of Linear Programming problem is called

What is meant by "Open Convex Region"?

Define: 'Objective function'.

Define feasible region in LPP.

Determine graphically the minimum value of the objective functionZ=50x+20y...(1)Z = 50x + 20y ... (1)subject to the constraints:2xy≥5...(2)2x y \geq 5 ... (2)3x+y≥3...(3)3x + y \geq 3 ... (3)2x−3y≤12...(4)2x -3y \leq 12 ... (4)x≥0,y≥0...(5)x \geq 0, y \geq 0 ... (5).

Minimize Z=7x+yZ = 7x + y subject to5x+y≥5,x+y≥3,x≥0,y≥05x + y\geq 5, x + y \geq 3, x \geq 0, y\geq 0.

Minimise Z = 3x + 2yZ = 3x + 2ysubject to the constraints:x + y \geq 8 ... (1)x + y \geq 8 ... (1)3x + 5y \leq 15 ... (2)3x + 5y \leq 15 ... (2)x \geq 0, y \geq 0 ... (3)x \geq 0, y \geq 0 ... (3).

Solve the following inequalities by graphical method:2x+y\ge 6,3x+4y\le 122x+y\ge 6,3x+4y\le 12

Solve the following linear programming problem graphically:Maximise Z = 4x + y ... (1)Z = 4x + y ... (1)subject to the constraints: x + y \leq 50 ... (2)x + y \leq 50 ... (2)3x + y\leq 90 ... (3)3x + y\leq 90 ... (3)x\geq 0, y\geq 0 ... (4)x\geq 0, y\geq 0 ... (4).

Find graphical solution for the following system of linear inequations:3x + 2y\leq 180; x + 2y \leq 120, x \geq 0, y\geq 03x + 2y\leq 180; x + 2y \leq 120, x \geq 0, y\geq 0Hence find co-ordinates of corner points of the common region.

Solve the following Linear Programming Problems graphically:Maximise Z = 3x + 2yZ = 3x + 2ysubject to x + 2y \leq 10, 3x + y \leq 15, x, y \geq 0x + 2y \leq 10, 3x + y \leq 15, x, y \geq 0.

Use graphical methodMaximize: z=2x+3yz=2x+3ySubject to:x+2y\le 40x+2y\le 406x+5y\le 1506x+5y\le 150x\ge 0, y\ge 0x\ge 0, y\ge 0

If possible, maximize z = x + 4yz = x + 4y, subject to 3x + 6y \leq 6, 4x + 8y \geq 163x + 6y \leq 6, 4x + 8y \geq 16 and x \geq 0, y \geq 0x \geq 0, y \geq 0.

Solve the following linear programming problem using graphical methodMinimize: Z=60x+30yZ=60x+30ySubject to:2x+3y\ge 1202x+3y\ge 1202x+y\ge 802x+y\ge 80x\ge 0, y\ge 0x\ge 0, y\ge 0

Find the maximum value of 4x+7y4x+7y with the conditions 3x+8y\leq 24, y\leq2, x\geq 03x+8y\leq 24, y\leq2, x\geq 0 and y\geq0y\geq0.

The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8Rs 8 and for the subsequence distance it is Rs 5 per KmRs 5 per Km. Taking the distance covered as x Kmx Km and total fare as Rs y.Rs y. write linear equation for this information, and draw its graph.

Solve each of the following linear programming problem by graphical method.Maximize Z=18x+10yZ=18x+10ySubject to 4x+y\le 204x+y\le 20 2x+3y\le 302x+3y\le 30 x,y\ge 0x,y\ge 0

Solve the following systems of inequalities graphically: x+y\leq 9, y > x, x\geq 1x+y\leq 9, y > x, x\geq 1

Solve 3x+2y\leq 12, x+2y\leq 16, x+y\leq 10, x\geq 0, y\geq 0.3x+2y\leq 12, x+2y\leq 16, x+y\leq 10, x\geq 0, y\geq 0.

Suppose f is the collection of all ordered pairs of real numbers and x=6 is the first element of some ordered pair in f. Suppose the vertical line through x=6 intersects the graph of f twice. Is f a function? Why or why not?

Draw the line passing through(2, 3) and (3,2). Find the sum of coordinates of the points at which this line meets the x-axis and y-axis.

Solve 4x + 3y > 124x + 3y > 12 graphically

Solve the following inequations \displaystyle \frac{2x+4}{x-1}\geq 5\displaystyle \frac{2x+4}{x-1}\geq 5

Solve & graph the solution set of -2 < 2x - 4-2 < 2x - 4 and \displaystyle -2x+5\geq 13,x\: \epsilon \: R\displaystyle -2x+5\geq 13,x\: \epsilon \: R

Solve the following system of linear inequality graphically\displaystyle 2x+y\geq 6 \displaystyle 2x+y\geq 6 ......(1)\displaystyle x-y\leq 2 \displaystyle x-y\leq 2 ......(2)

If SS is the solution region of the inequalities \begin{cases} x+3y\le 18 \\ 2x-3y\le 9 \end{cases}\begin{cases} x+3y\le 18 \\ 2x-3y\le 9 \end{cases} and (a,b) \in S(a,b) \in S, then the minimum possible value for bb is

Solve the following systems of inequalities graphically: x+y\leq 9, y>x, x\geq 1x+y\leq 9, y>x, x\geq 1.

Maximize f=4x-yf=4x-y, subject to the constraints7x+4y\le 287x+4y\le 28, 2y\le 72y\le 7, x\ge 0x\ge 0, y\ge 0y\ge 0.

Maximize f=3x+yf=3x+y subject to the constraints8x+5y\le 408x+5y\le 40; 4x+3y\ge 124x+3y\ge 12; x\ge 0x\ge 0 and y\ge 0y\ge 0.

A firm has the cost function C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50 and demand function x = 100 - px = 100 - p.Formulate the total profit function PP in terms of xx

By graphical method solve the following linear programming problem for maximization.Objective function Z = 1000 x + 600 yConstraints x + y \leq 200x + y \leq 2004x - y \leq 04x - y \leq 0x \geq 20, x \geq 0, y \geq 0x \geq 20, x \geq 0, y \geq 0.

A firm has the cost function C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50 and demand function x = 100 - px = 100 - p.Write the total revenue function in terms of xx

A shopkeeper sells not more than 3030 shirts of each colour. Atleast twice as many white ones are sold as green ones. If the profit on each of the white be Rs. 2020 and that of green be Rs. 2525; then find out how many of each kind be sold to give him a maximum profit. (Graph need not be drawn)

Solve the following L.P.P. by graphical method:Maximise: Z = 6x + 4yZ = 6x + 4ySubject to x\leq 2, x + y \leq 3, -2x + y \leq 1, x \geq 0, y \geq 0x\leq 2, x + y \leq 3, -2x + y \leq 1, x \geq 0, y \geq 0

Solve the following linear programming problem graphically:Maximize Z = 7x + 10yZ = 7x + 10ysubject to the constraints4x + 6y \leq 2404x + 6y \leq 2406x + 3y \leq 2406x + 3y \leq 240x \geq 10x \geq 10x \geq 0, y \geq 0x \geq 0, y \geq 0

Consider the linear inequalities2x+3y\le 6, 2x+y\le 4, x\ge 0, y\ge 02x+3y\le 6, 2x+y\le 4, x\ge 0, y\ge 0(a) Mark the feasible region.(b) Maximise the function z=4x+5yz=4x+5y subject to the given constraints.

Solve graphically the linear inequalities2x+3y\le 7,x+2y\le 42x+3y\le 7,x+2y\le 4x\ge 0,y\ge 0 x\ge 0,y\ge 0 If z=6x+5yz=6x+5y is the objective function, find its maximum value.

Define feasible region.

Consider the following L.P.P.Maximize Z=3x+2yZ=3x+2ySubject to the constraintsx+2y\le 10x+2y\le 103x+y\le 153x+y\le 15x,y\ge 0x,y\ge 0(a) Draw its feasible region.(b) Find the corner points of the feasible region.(c) Find the maximum value of ZZ.

By Graphical method solve the following L.P.P under the following constraints.x+y\le 24,2x+y\le 32,x,y\ge 0x+y\le 24,2x+y\le 32,x,y\ge 0. Find the maximum value of z=150x+250yz=150x+250y

Solve the linear programming problem by graphical method with the following restrictions.3x+5y \leq 5; 5x+2y\leq 10; x\geq 0, y\geq 03x+5y \leq 5; 5x+2y\leq 10; x\geq 0, y\geq 0 and maximize Z=5x+3yZ=5x+3y.

Show the solution of the problem of linear programming under the following restrictions by graphical method: x+y\leq 4, x \geq 0, y\geq 0x+y\leq 4, x \geq 0, y\geq 0.

Solve the following Linear Programming Problems graphically:Minimise Z = -3x + 4yZ = -3x + 4ysubject to x + 2y\leq 8, 3x + 2y \leq 12, x\geq 0, y \geq 0x + 2y\leq 8, 3x + 2y \leq 12, x\geq 0, y \geq 0.

Solve the following Linear Programming Problems graphically:Maximise Z = 5x + 3yZ = 5x + 3ysubject to 3x + 5y\leq 15, 5x + 2y \leq 10, x \geq 0, y \geq 03x + 5y\leq 15, 5x + 2y \leq 10, x \geq 0, y \geq 0.

Solve the following Linear Programming Problems graphically :Minimize Z = 3x + 5yZ = 3x + 5ysubject to x + y \leq 4, x \geq 0x + y \leq 4, x \geq 0 and y \ge 0y \ge 0

Solve the following linear programming problem graphically :Minimise Z = 200 x + 500 y ... (1)Z = 200 x + 500 y ... (1)subject to the constraints:x + 2y \geq 10 ... (2)x + 2y \geq 10 ... (2)3x + 4y \leq 24 ... (3)3x + 4y \leq 24 ... (3)x \geq 0, y \geq 0 ... (4)x \geq 0, y \geq 0 ... (4).

If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

Solve using graphical method:Maximize: Z=4x+5yZ=4x+5ySubject to:2x+5y\le 252x+5y\le 256x+5y\le 456x+5y\le 454949x\ge 0, y\ge 0x\ge 0, y\ge 0

Solve the following problem:Maximize Z=-P{x}_{1}+4{x}_{2}Z=-P{x}_{1}+4{x}_{2}Subject to: -3{x}_{1}+{x}_{2}\le 6-3{x}_{1}+{x}_{2}\le 6{x}_{1}+2{x}_{2}\le 4{x}_{1}+2{x}_{2}\le 4{x}_{2}\le -3{x}_{2}\le -3No lower bound constraint for {x}_{1}{x}_{1}

Using graphical methodFind the corner points for2x+5y\le 252x+5y\le 256x+5y\le 456x+5y\le 45x\ge 0,y\ge 0x\ge 0,y\ge 0

Use graphical methodMinimize: z=600x+900yz=600x+900ySubject to:40x+60y\ge 48040x+60y\ge 48030x+15y\ge 18030x+15y\ge 180x\ge 0,y\ge 0x\ge 0,y\ge 0

Construct the graphs of the following functions.\displaystyle y \, = \, x \, | x | \, - \, 4x \, - \, 5.\displaystyle y \, = \, x \, | x | \, - \, 4x \, - \, 5.

Construct the graphs of the following functions.\displaystyle y \, = \, (2x \, + \, 1) \, / \, (x \, - \, 1).\displaystyle y \, = \, (2x \, + \, 1) \, / \, (x \, - \, 1).

Construct the graphs of the following functions.\displaystyle y \, = \, ( |1 \, - \, x \, | \, + \, 2) \, (x \, + \, 1).\displaystyle y \, = \, ( |1 \, - \, x \, | \, + \, 2) \, (x \, + \, 1).

Construct the graphs of the following functions.\displaystyle y \, = \, | 2x^2 \, - \, 3x \, + \, | x \, - \, 1| \, |.\displaystyle y \, = \, | 2x^2 \, - \, 3x \, + \, | x \, - \, 1| \, |.

Construct the graphs of the following functions.\displaystyle y \, = \, 1 / | x |.\displaystyle y \, = \, 1 / | x |.

Construct the graphs of the following functions.\displaystyle y \, = \, \frac{| x \, - \, 1 |}{x \, - \, 1} \, (x^2 \, + \, 3).\displaystyle y \, = \, \frac{| x \, - \, 1 |}{x \, - \, 1} \, (x^2 \, + \, 3).

Construct the graphs of the following functions.\displaystyle y \, = \, x (x \, - \, 3)^2.\displaystyle y \, = \, x (x \, - \, 3)^2.

Solve the following maximal assignment problem:Branch ManagerMonthly Business(Rs. lakh)AABBCCDDPP111111119999QQ1313161611111010RR12121717131388SS1616141416161212

Minimize and maximize Z=600x+400yZ=600x+400ySubject to the constraintsx+2y\le 12x+2y\le 122x+y\le 122x+y\le 124x+5y\le 204x+5y\le 20x\ge 0;y\ge 0x\ge 0;y\ge 0 by graphical method

Solve the following using graphical method:Minimize: Z = 3x + 5yZ = 3x + 5y subject to 2x+ 3y \geq 12,2x+ 3y \geq 12, -x+y \leq 3-x+y \leq 3, x \leq 4, \,\, y \geq 3, \,\, x \geq 0, \,\, y \geq 0x \leq 4, \,\, y \geq 3, \,\, x \geq 0, \,\, y \geq 0

Solve the following minimal assignment problem and hence find minimum time where '-''-' indicates that job cannot be assigned to the machine:MachinesProcessing time in hoursAABBCCDDEEM_{1}M_{1}991111151510101111M_{2}M_{2}121299-101099M_{3}M_{3}-11111414111177M_{4}M_{4}14148812127788

Solve the following problem graphically:Minimise and Maximisez = 3x + 9yz = 3x + 9ySubject to the constraints:x + 3y \leq 60x + 3y \leq 60x + y \geq 10x + y \geq 10x \leq yx \leq yx\geq 0, y\geq 0x\geq 0, y\geq 0.

Construct the graphs of the following functions.\displaystyle y \, = \, (2 | x | -1) \, / \, (x \, - \, 3).\displaystyle y \, = \, (2 | x | -1) \, / \, (x \, - \, 3).

In solving LP problem : minimize z = 6x + 10yz = 6x + 10y, subject to x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0, x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0, which constraints are reduntant?

Minimise Z= 3x +2yZ= 3x +2ysubject to constraints: x+y \geq 8x+y \geq 83x+ 5y\leq 153x+ 5y\leq 15x\geq 0, y \geq 0x\geq 0, y \geq 0

One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 1515 camels were seen on the bank of a river. Find the total number of camels.

Solve the following system of inequalities graphically 2x+y\ge 6, 3x+4y\le 122x+y\ge 6, 3x+4y\le 12

Maximize Z=x+2yZ=x+2y subject to 2x+y\ge 3,x+2y\ge 6, x,y\ge 02x+y\ge 3,x+2y\ge 6, x,y\ge 0.

Draw the line 2y=3x+42y=3x+4 on the xyxy plane.

Solve the following system of inequalities graphically x+y\ge 4, 2x-y> 12x+y\ge 4, 2x-y> 12

Solve the following system of inequalities graphically x+y\le 9, y>x, x\ge 0x+y\le 9, y>x, x\ge 0

Solve the following system of inequalities graphically 5x+4y\le 20, x\ge 1, y\ge 25x+4y\le 20, x\ge 1, y\ge 2

Solve the following system of inequalities graphically 2x-y>1, x-2y<-12x-y>1, x-2y<-1

Solve the following system of inequalities graphically 2x+y\ge 8, x+2y\ge 102x+y\ge 8, x+2y\ge 10

Solve the following equation
$\displaystyle 2\left ( 2x+3 \right )-10\leq 6\left ( x-2 \right )$

Check, whether the half plane $3x+6y \geq 0$ contains $(1, 1)$ , if so shade the plane containing $(1, 1)$ .

Shade the following half plane: $5x+10y-20\geq 0$ .

Check, whether the half plane $x+y+4 > 0$ contains the origin. Identify the plane and shade it.

Determine graphically the minimum value of the objective function
$Z = 50x + 20y ... (1)$
subject to the constraints:
$2x y \geq 5 ... (2)$
$3x + y \geq 3 ... (3)$
$2x -3y \leq 12 ... (4)$
$x \geq 0, y \geq 0 ... (5)$ .

Minimize $Z = 7x + y$ subject to
$5x + y\geq 5, x + y \geq 3, x \geq 0, y\geq 0$ .

Minimise $Z = 3x + 2y$
subject to the constraints:
$x + y \geq 8 ... (1)$
$3x + 5y \leq 15 ... (2)$
$x \geq 0, y \geq 0 ... (3)$ .

Solve the following inequalities by graphical method:
$2x+y\ge 6,3x+4y\le 12$

Solve the following linear programming problem graphically:
Maximise $Z = 4x + y ... (1)$
subject to the constraints: $x + y \leq 50 ... (2)$
$3x + y\leq 90 ... (3)$
$x\geq 0, y\geq 0 ... (4)$ .

Find graphical solution for the following system of linear inequations:
$3x + 2y\leq 180; x + 2y \leq 120, x \geq 0, y\geq 0$
Hence find co-ordinates of corner points of the common region.

Solve the following Linear Programming Problems graphically:
Maximise $Z = 3x + 2y$
subject to $x + 2y \leq 10, 3x + y \leq 15, x, y \geq 0$ .

Use graphical method
Maximize: $z=2x+3y$
Subject to:
$x+2y\le 40$
$6x+5y\le 150$
$x\ge 0, y\ge 0$

If possible, maximize $z = x + 4y$ , subject to $3x + 6y \leq 6, 4x + 8y \geq 16$ and $x \geq 0, y \geq 0$ .

Solve the following linear programming problem using graphical method
Minimize: $Z=60x+30y$
Subject to:
$2x+3y\ge 120$
$2x+y\ge 80$
$x\ge 0, y\ge 0$

Find the maximum value of $4x+7y$ with the conditions $3x+8y\leq 24, y\leq2, x\geq 0$ and $y\geq0$ .

The taxi fare in a city is as follows: For the first kilometre, the fare is $Rs 8$ and for the subsequence distance it is $Rs 5 per Km$ . Taking the distance covered as $x Km$ and total fare as $Rs y.$ write linear equation for this information, and draw its graph.

Solve each of the following linear programming problem by graphical method.
Maximize $Z=18x+10y$
Subject to $4x+y\le 20$
$2x+3y\le 30$
$x,y\ge 0$

Solve the following systems of inequalities graphically: $x+y\leq 9, y > x, x\geq 1$

Solve $3x+2y\leq 12, x+2y\leq 16, x+y\leq 10, x\geq 0, y\geq 0.$

Solve $4x + 3y > 12$ graphically

Solve the following inequations $\displaystyle \frac{2x+4}{x-1}\geq 5$

Solve & graph the solution set of $-2 < 2x - 4$ and $\displaystyle -2x+5\geq 13,x\: \epsilon \: R$

Solve the following system of linear inequality graphically
$\displaystyle 2x+y\geq 6$ ......(1)
$\displaystyle x-y\leq 2$ ......(2)

If $S$ is the solution region of the inequalities $\begin{cases} x+3y\le 18 \\ 2x-3y\le 9 \end{cases}$ and $(a,b) \in S$ , then the minimum possible value for $b$ is

Solve the following systems of inequalities graphically: $x+y\leq 9, y>x, x\geq 1$ .

Maximize $f=4x-y$ , subject to the constraints
$7x+4y\le 28$ , $2y\le 7$ , $x\ge 0$ , $y\ge 0$ .

Maximize $f=3x+y$ subject to the constraints
$8x+5y\le 40$ ; $4x+3y\ge 12$ ; $x\ge 0$ and $y\ge 0$ .

A firm has the cost function $C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50$ and demand function $x = 100 - p$ .
Formulate the total profit function $P$ in terms of $x$

By graphical method solve the following linear programming problem for maximization.
Objective function Z = 1000 x + 600 y
Constraints $x + y \leq 200$
$4x - y \leq 0$
$x \geq 20, x \geq 0, y \geq 0$ .

A firm has the cost function $C = \dfrac {x^{3}}{3} - 7x^{2} + 111x + 50$ and demand function $x = 100 - p$ .
Write the total revenue function in terms of $x$

A shopkeeper sells not more than $30$ shirts of each colour. Atleast twice as many white ones are sold as green ones. If the profit on each of the white be Rs. $20$ and that of green be Rs. $25$ ; then find out how many of each kind be sold to give him a maximum profit. (Graph need not be drawn)

Solve the following L.P.P. by graphical method:

Maximise: $Z = 6x + 4y$

Subject to $x\leq 2, x + y \leq 3, -2x + y \leq 1, x \geq 0, y \geq 0$

Solve the following linear programming problem graphically:
Maximize $Z = 7x + 10y$
subject to the constraints
$4x + 6y \leq 240$
$6x + 3y \leq 240$
$x \geq 10$
$x \geq 0, y \geq 0$

Consider the linear inequalities
$2x+3y\le 6, 2x+y\le 4, x\ge 0, y\ge 0$
(a) Mark the feasible region.
(b) Maximise the function $z=4x+5y$ subject to the given constraints.

Solve graphically the linear inequalities
$2x+3y\le 7,x+2y\le 4$
$x\ge 0,y\ge 0$
If $z=6x+5y$ is the objective function, find its maximum value.

Consider the following L.P.P.
Maximize $Z=3x+2y$
Subject to the constraints
$x+2y\le 10$
$3x+y\le 15$
$x,y\ge 0$
(a) Draw its feasible region.
(b) Find the corner points of the feasible region.
(c) Find the maximum value of $Z$ .

By Graphical method solve the following L.P.P under the following constraints.
$x+y\le 24,2x+y\le 32,x,y\ge 0$ . Find the maximum value of $z=150x+250y$

Solve the linear programming problem by graphical method with the following restrictions.
$3x+5y \leq 5; 5x+2y\leq 10; x\geq 0, y\geq 0$ and maximize $Z=5x+3y$ .

Show the solution of the problem of linear programming under the following restrictions by graphical method: $x+y\leq 4, x \geq 0, y\geq 0$ .

Solve the following Linear Programming Problems graphically:
Minimise $Z = -3x + 4y$
subject to $x + 2y\leq 8, 3x + 2y \leq 12, x\geq 0, y \geq 0$ .

Solve the following Linear Programming Problems graphically:
Maximise $Z = 5x + 3y$
subject to $3x + 5y\leq 15, 5x + 2y \leq 10, x \geq 0, y \geq 0$ .

Solve the following Linear Programming Problems graphically :
Minimize $Z = 3x + 5y$
subject to $x + y \leq 4, x \geq 0$ and $y \ge 0$

Solve the following linear programming problem graphically :
Minimise $Z = 200 x + 500 y ... (1)$
subject to the constraints:
$x + 2y \geq 10 ... (2)$
$3x + 4y \leq 24 ... (3)$
$x \geq 0, y \geq 0 ... (4)$ .

Solve using graphical method:
Maximize: $Z=4x+5y$
Subject to:
$2x+5y\le 25$
$6x+5y\le 45$
$49$
$x\ge 0, y\ge 0$

Solve the following problem:
Maximize $Z=-P{x}_{1}+4{x}_{2}$
Subject to: $-3{x}_{1}+{x}_{2}\le 6$
${x}_{1}+2{x}_{2}\le 4$
${x}_{2}\le -3$
No lower bound constraint for ${x}_{1}$

Using graphical method
Find the corner points for
$2x+5y\le 25$
$6x+5y\le 45$
$x\ge 0,y\ge 0$

Use graphical method
Minimize: $z=600x+900y$
Subject to:
$40x+60y\ge 480$
$30x+15y\ge 180$
$x\ge 0,y\ge 0$

Construct the graphs of the following functions.
$\displaystyle y \, = \, x \, | x | \, - \, 4x \, - \, 5.$

Construct the graphs of the following functions.
$\displaystyle y \, = \, (2x \, + \, 1) \, / \, (x \, - \, 1).$

Construct the graphs of the following functions.
$\displaystyle y \, = \, ( |1 \, - \, x \, | \, + \, 2) \, (x \, + \, 1).$

Construct the graphs of the following functions.
$\displaystyle y \, = \, | 2x^2 \, - \, 3x \, + \, | x \, - \, 1| \, |.$

Construct the graphs of the following functions.
$\displaystyle y \, = \, 1 / | x |.$

Construct the graphs of the following functions.
$\displaystyle y \, = \, \frac{| x \, - \, 1 |}{x \, - \, 1} \, (x^2 \, + \, 3).$

Construct the graphs of the following functions.
$\displaystyle y \, = \, x (x \, - \, 3)^2.$

Solve the following maximal assignment problem:
Branch Manager Monthly Business
(Rs. lakh)
$A$ $B$ $C$ $D$
$P$ $11$ $11$ $9$ $9$
$Q$ $13$ $16$ $11$ $10$
$R$ $12$ $17$ $13$ $8$
$S$ $16$ $14$ $16$ $12$

Minimize and maximize $Z=600x+400y$
Subject to the constraints
$x+2y\le 12$
$2x+y\le 12$
$4x+5y\le 20$
$x\ge 0;y\ge 0$ by graphical method

Solve the following using graphical method:
Minimize: $Z = 3x + 5y$ subject to $2x+ 3y \geq 12,$ $-x+y \leq 3$ , $x \leq 4, \,\, y \geq 3, \,\, x \geq 0, \,\, y \geq 0$

Solve the following problem graphically:
Minimise and Maximise
$z = 3x + 9y$
Subject to the constraints:
$x + 3y \leq 60$
$x + y \geq 10$
$x \leq y$
$x\geq 0, y\geq 0$ .

Construct the graphs of the following functions.
$\displaystyle y \, = \, (2 | x | -1) \, / \, (x \, - \, 3).$

In solving LP problem : minimize $z = 6x + 10y$ , subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0,$ which constraints are reduntant?

Minimise $Z= 3x +2y$
subject to constraints:
$x+y \geq 8$
$3x+ 5y\leq 15$
$x\geq 0, y \geq 0$

One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining $15$ camels were seen on the bank of a river. Find the total number of camels.

Solve the following system of inequalities graphically $2x+y\ge 6, 3x+4y\le 12$

Maximize $Z=x+2y$ subject to $2x+y\ge 3,x+2y\ge 6, x,y\ge 0$ .

Draw the line $2y=3x+4$ on the $xy$ plane.

Solve the following system of inequalities graphically $x+y\ge 4, 2x-y> 12$

Solve the following system of inequalities graphically $x+y\le 9, y>x, x\ge 0$

Solve the following system of inequalities graphically $5x+4y\le 20, x\ge 1, y\ge 2$

Solve the following system of inequalities graphically $2x-y>1, x-2y<-1$

Solve the following system of inequalities graphically $2x+y\ge 8, x+2y\ge 10$

Solve the following system of inequalities graphically $x+y\le 6, x+y\ge 4$

Solve the following system of inequalities graphically $3x+4y\le 60, x+3y\le 30, x\ge 0, y\ge 0$

If the point $(2K-3,K+2)$ lies on the graph of equation $2x+3y+15=0$ , find the value of k.

Solve the following Linear programming problems graphically :
Maximize $z = 3x + 4y$ subject to the constraints : $x + y \le 4, \, x \ge 0 , \, y \ge 0$

See the given figure and write the following:
$(i)$ The coordinates of $B$
$(ii)$ The coordinates of $C$
$(iii)$ The point identified by the coordinates $(-3,-5)$
$(iv)$ The ordinate of the point $B$ .

Maximize $Z=-x+2y$ , subject to the constraints: $x\geq 3; x+y\geq 5, x+2 y\geq 6, y\geq 0$ .

Solve the following system of inequalities graphically $2x+y\ge 4, x+y\le 3, 2x-3y\le 6$

$x-y\leq -1,x+y\leq 0,x,y\geq0$

Find the solution of the following LP problem ''Maximize z=x+ y subject $x + y \leq 1,-3x+y\geq 3,x\geq 0,y\geq 0''$ is.............

Maximize $z=3x+5y$ , Subject to $x+4y \le 24$ , $3x+y \le 21,x+y \le 9,x \ge 0$ . Also find maximum value of $z$ .

	$A$	$B$
Grinding	$1$	$2$
Turning	$3$	$1$
Assembling	$6$	$3$
Testing	$5$	$4$

	$kg$ per bag
	Brand $P$	Brand $Q$
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$

	Area occupied by the machine	Labour force for each machine	Daily output in units
Machine $A$	$1000.sq.\ m$	$12\ men$	$60$
Machine $B$	$1200\ sq.\ m$	$8\ men$	$40$

	Compounds		Minimum requirement
	$A$	$B$
Ingredient $C$	$1$	$2$	$80$
Ingredient $D$	$3$	$1$	$75$
Cost (in $Rs$ ) $per\ kg$	$4$	$6$

	Transportation cost per quintal (in Rs)
To $\downarrow$ \ From $\to$	$A$	$B$
$D$	$6.00$	$4.00$
$E$	$3.00$	$2.00$
$F$	$2.50$	$3.00$