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CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Linear Programming
Quiz 4
The maximum profit that can be made in a day is:
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Rs. $$450$$
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Rs. $$800$$
0%
Rs. $$650$$
0%
Rs. $$525$$
Explanation
let the number of normal calculators produced in a day be $$x$$ and
the number of scientific calculators produced in a day be $$y$$
the minimum of total calculators to be produced per day is $$200\implies x+y\geq 200$$
Given, the minimum number of
normal calculators to be produced per day is $$100 \implies x\geq 100$$ and
the minimum number of scientific
calculators to be produced per day is $$80 \implies y\geq 80$$
Also g
iven, the maximum number of
normal calculators can be produced per day is $$200 \implies x\leq 200$$ and
the maximum number of scientific
calculators can be produced per day is $$170 \implies y\leq 170$$
A normal calculator incurred a loss of $$Rs.2$$
For $$x$$ normal calculators, the loss is $$Rs.2x$$
A scientific calculator gained a profit of $$Rs.5$$
For $$xy$$ scientific calculators, the gain is $$Rs.5y$$
Therefore, profit of the manufacturer $$P=5y-2x$$
In the above figure, the blue shaded region is the feasible region with five corner points.
$$(100,170), (200,170), (200,80), (120,80), (100,100)$$
$$(100,170)$$ is the point where
$$x=100$$
intersects $$y=170$$
$$(200,170)$$ is the point where
$$x=200$$
intersects $$y=170$$
$$(200,80)$$ is the point where
$$x=200$$
intersects $$y=80$$
$$(120,80)$$ is the point where
$$x+y=200$$
intersects $$y=80$$
I.e., substituting $$y=80 \implies x+80 =200 \implies x=200-80 \implies x=120$$
$$(100,100)$$ is the point where
$$x+y=200$$
intersects $$x=100$$
I.e., substituting $$x=100 \implies 100+y =200 \implies x=200-100 \implies x=100$$
Now substituting the corner points the profit equation,
substituting $$(100,170) \implies P=5\times 170-2\times 100=650$$
substituting $$(200,170) \implies P=5\times 170-2\times 200=450$$
substituting $$(200,80) \implies P=5\times 80-2\times 200=0$$
substituting $$(120,80) \implies P=5\times 80-2\times 120=160$$
substituting $$(100,100) \implies P=5\times 100-2\times 100=300$$
$$650 $$ is the maximum profit
The Convex Polygon Theorem states that the optimum (maximum or minimum) solution of a LPP is attained at atleast one of the ______ of the convex set over which the solution is feasible.
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origin
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corner points
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centre
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edge
Explanation
The fundamental theorem of programming (i.e., Convex Polygon Theorem) states that the optimum value(maximum or minimum) of a linear programming problem over a convex region occur at the corner points.
In order to obtain maximum profit, the quantity of normal and scientific calculators to be manufactured daily is:
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$$(100,170)$$
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$$(200,170)$$
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$$(100,80)$$
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$$(200,80)$$
Explanation
let the number of normal calculators produced in a day be $$x$$ and
the number of scientific calculators produced in a day be $$y$$
the minimum of total calculators to be produced per day is $$200\implies x+y\geq 200$$
Given, the minimum number of
normal calculators to be produced per day is $$100 \implies x\geq 100$$ and
the minimum number of scientific
calculators to be produced per day is $$80 \implies y\geq 80$$
Also g
iven, the maximum number of
normal calculators can be produced per day is $$200 \implies x\leq 200$$ and
the maximum number of scientific
calculators can be produced per day is $$170 \implies y\leq 170$$
A normal calculator incurred a loss of $$Rs.2$$
For $$x$$ normal calculators, the loss is $$Rs.2x$$
A scientific calculator gained a profit of $$Rs.5$$
For $$xy$$ scientific calculators, the gain is $$Rs.5y$$
Therefore, profit of the manufacturer $$P=5y-2x$$
substituting the options in the equation
$$P=5y-2x$$
and finding the maximum value and verifying whether the option satisfies the given constraints.
substituting option A i.e., $$(x,y)=(100,170)$$
$$x+y\geq 200\implies 100+170=270\geq 200$$ True
$$x\geq 100 $$ and $$x\leq 200$$ True
$$y\geq 80 $$ and $$x\leq 170$$ True
Hence,
$$P=5y-2x = 5*170-2*100=650$$
substituting option B i.e., $$(x,y)=(200,170)$$
$$x+y\geq 200\implies 200+170=370\geq 200$$ True
$$x\geq 100 $$ and $$x\leq 200$$ True
$$y\geq 80 $$ and $$x\leq 170$$ True
Hence,
$$P=5y-2x = 5*170-2*200=450$$
substituting option C i.e., $$(x,y)=(100,80)$$
$$x+y\geq 200\implies 100+80=180\geq 200$$ False
substituting option D i.e., $$(x,y)=(200,80)$$
$$x+y\geq 200\implies 200+80=280\geq 200$$ True
$$x\geq 100 $$ and $$x\leq 200$$ True
$$y\geq 80 $$ and $$x\leq 170$$ True
Hence,
$$P=5y-2x = 5*80-2*200=0$$
Therefore, $$(100,170)$$ satisfies all constraints and produces the maximum value than the other options.
The region on the graph sheet with satisfies the constraints including the non- negativity restrictions is called the _______ space.
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solution
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interval
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concave
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convex
Explanation
In linear programming, the feasible region is defined as the region in which all the set of points which satisfy the given constraints and the non-negativity restrictions.
In the above figure, the blue shaded region is the feasible region which contains the solution set.
In order to maximize the profit of the company, the optimal solution of which of the following equations is required?
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$$P = x+y-200$$
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$$P = 5y-2x$$
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$$P = y - 80$$
0%
$$P = 200-x$$
Explanation
let the number of normal calculators produced in a day be $$x$$ and
the number of scientific calculators produced in a day be $$y$$
the minimum of total calculators to be produced per day is $$200\implies x+y\geq 200$$
Given, the minimum number of
normal calculators to be produced per day is $$100 \implies x\geq 100$$ and
the minimum number of scientific
calculators to be produced per day is $$80 \implies y\geq 80$$
Also g
iven, the maximum number of
normal calculators can be produced per day is $$200 \implies x\leq 200$$ and
the maximum number of scientific
calculators can be produced per day is $$170 \implies y\leq 170$$
A normal calculator incurred a loss of $$Rs.2$$
For $$x$$ normal calculators, the loss is $$Rs.2x$$
A scientific calculator gained a profit of $$Rs.5$$
For $$xy$$ scientific calculators, the gain is $$Rs.5y$$
Therefore, profit of the manufacturer $$P=5y-2x$$
How many acres of each (wheat and rye) should the farmer plant in order to get maximum profit?
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$$(5,5)$$
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$$(4,4)$$
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$$(4,5)$$
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$$(4,3)$$
Explanation
let $$x$$ be the acres of wheat planted and
$$y$$ be the acres of rye planted
Given that there are a total of $$10$$ acres of land to plant.
Atleast $$7$$ acres is to be planted i.e., $$x+y\geq 7$$
Given that the cost to plant one acre of wheat is $$\$200$$
Therefore, the cost for $$x$$ acres of wheat is $$200x$$
Given that the cost to plant one acre of rye is $$\$100$$
Therefore, the cost for $$y$$ acres of rye is $$100y$$
Given that, amount for planting wheat and rye is $$\$1200$$
Therefore the total cost to plant wheat and rye is $$200x+100y\leq 1200\implies 2x+y\leq 12$$
Given that, the time taken to plant one acre of wheat is $$1$$ hr
Therefore, the time taken to plant $$x$$ acres of wheat is $$x$$ hrs
Given that, the time taken to plant one acre of rye is $$2$$ hrs
Therefore, the time taken to plant $$y$$ acres of rye is $$2y$$ hrs
Given that, the total time for planting is $$12$$ hrs
Therefore, the total time to plant wheat and rye is $$x+2y\leq 12$$
Given that, one acre of wheat yields a profit of $$\$500$$
Therefore, the profit from $$x$$ acres of wheat is $$500x$$
Given that, one acre of rye yields a profit of $$\$300$$
Therefore, the profit from $$y$$ acres of wheat is $$300y$$
therefore the total profit from the wheat and rye is $$P=500x+300y$$
Now substituting the options in the profit expression and verifying
Substituting option A $$(x,y)=(5,5)$$
$$x+y\geq 0\implies 5+5\geq7\implies 10\geq 7$$ True
$$2x+y\leq 12\implies 2(5)+5\leq12\implies 15\leq 12$$ False
Substituting option B $$(x,y)=(4,4)$$
$$x+y\geq 0\implies 4+4\geq7\implies 8\geq 7$$ True
$$2x+y\leq 12\implies 2(4)+4\leq12\implies 12\leq 12$$ True
$$x+2y\leq 12\implies 4+2(4)\leq12\implies 12\leq 12$$ True
Substituting option C $$(x,y)=(4,5)$$
$$x+y\geq 0\implies 4+5\geq7\implies 9\geq 7$$ True
$$2x+y\leq 12\implies 2(4)+5\leq12\implies 13\leq 12$$ False
Substituting option D $$(x,y)=(4,3)$$
$$x+y\geq 0\implies 4+3\geq7\implies 7\geq 7$$ True
$$2x+y\leq 12\implies 2(4)+3\leq12\implies 11\leq 12$$ True
$$x+2y\leq 12\implies 4+2(3)\leq12\implies 10\leq 12$$ True
$$(4,4), (4,3)$$ satisfies the constraints. Therefore finding the profit
for $$(4,4)$$, $$P=500x+300y=500(4)+300(4)=3200$$
for $$(4,3)$$, $$P=500x+300y=500(4)+300(3)=2900$$
Therefore the maximum profit is attained at $$(4,4)$$
How many acres of land will be left unplanted if the farmer aims for a maximum profit?
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
let $$x$$ be the acres of wheat planted and
$$y$$ be the acres of rye planted
Given that there are a total of $$10$$ acres of land to plant.
Atleast $$7$$ acres is to be planted i.e., $$x+y\geq 7$$
Given that the cost to plant one acre of wheat is $$\$200$$
Therefore, the cost for $$x$$ acres of wheat is $$200x$$
Given that the cost to plant one acre of rye is $$\$100$$
Therefore, the cost for $$y$$ acres of rye is $$100y$$
Given that, an amount for planting wheat and rye is $$\$1200$$
Therefore the total cost to plant wheat and rye is $$200x+100y\leq 1200\implies 2x+y\leq 12$$
Given that, the time taken to plant one acre of wheat is $$1$$ hr
Therefore, the time taken to plant $$x$$ acres of wheat is $$x$$ hrs
Given that, the time taken to plant one acre of rye is $$2$$ hrs
Therefore, the time taken to plant $$y$$ acres of rye is $$2y$$ hrs
Given that, the total time for planting is $$12$$ hrs
Therefore, the total time to plant wheat and rye is $$x+2y\leq 12$$
Given that, one acre of wheat yields a profit of $$\$500$$
Therefore, the profit from $$x$$ acres of wheat is $$500x$$
Given that, one acre of rye yields a profit of $$\$300$$
Therefore, the profit from $$y$$ acres of wheat is $$300y$$
therefore the total profit from the wheat and rye is $$P=500x+300y$$
In the above figure, the blue shaded region is the feasible region with three corner points.
$$(4,4), (2,5), (5,2)$$
Now substituting the corner points the profit equation,
substituting
$$(4,4) \implies P=500x+300y=500(4)+300(4)=3200$$
substituting
$$(2,5) \implies P=500x+300y=500(2)+300(5)=2500$$
substituting
$$(5,2) \implies P=500x+300y=500(5)+300(2)=3100$$
$$\$3200 $$ is the maximum profit is attained by planting $$4$$ acres of wheat and $$4$$ acres of rye.
i.e., a total of $$4+4=8$$ acres of land is planted. But, the total land is $$10$$ acres. therefore, the unplanted land is $$10-8=2$$ acres
If $$a,b,c \in +R$$ such that $$\lambda abc$$ is the minimum value of $$a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)$$, then $$\lambda=$$
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$$1$$
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$$3$$
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$$4$$
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None of the above.
Explanation
We know that $$A.M.\geq G.M.$$
Therefore, $$\dfrac{b^2+c^2}{2}\geq \sqrt{b^2c^2}$$
$$\implies b^2+c^2\geq 2bc$$
Multiplying $$a$$ on both sides doesn’t change the inequality. Since, given that $$a$$ is positive.
$$\implies a(b^2+c^2)\geq 2abc$$ ————(1)
Similarly, $$b(a^2+c^2)\geq 2abc$$ ————(2)
and $$c(a^2+b^2)\geq 2abc$$ —————(3)
adding (1), (2) and (3) we get
$$a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)\geq 2abc+2abc+2abc$$
$$\implies a(b^2+c^2)+b(a^2+b^2)+c(a^2+b^2)\geq 6abc$$
Therefore $$\lambda$$ is $$6$$
Consider the objective function $$Z = 40x + 50y$$ The minimum number of constraints that are required to maximize $$Z$$ are
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$$4$$
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$$2$$
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$$3$$
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$$1$$
Explanation
Since in the given function $$Z=40x+50y$$, two variables are used.
So, the two constraints will be $$x \geq 0, y \geq 0$$ and the third one will be of the type $$ax + by \geq c$$.
Hence, at least $$3$$ constraints are required.
Maximum value of $$z=3x+4y$$ subject to $$x-y\le -1,-x+y\le 0,x,y\ge 0$$ is given by?
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$$1$$
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$$4$$
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$$6$$
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None of the these
Explanation
No common feasible area
No commonpoint,
So, $${ Z }_{ max }$$ cant find for these curves.
Hence none of these.
The objective function $$z={ x }_{ 1 }+{ x }_{ 2 }$$, subject to $${ x }_{ 1 }+{ x }_{ 2 }\le 10,{ -2x }_{ 1 }+3{ x }_{ 2 }\le 15,{ x }_{ 1 }\le 6$$, $${ x }_{ 1 },{ x }_{ 2 }\ge 0$$ has maximum value .................. of the feasible region.
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at only one point
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at only two points
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at every point of the segment joining two points
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at every point of the line joining two points
Explanation
For the equation, $$x_1+x_2=10$$,
$$x$$
$$0$$
$$10$$
$$5$$
$$y$$
$$10$$
$$0$$
$$5$$
For the equation, $$-2x_1+3x_2=15$$,
$$x$$
$$0$$
$$-3$$
$$3$$
$$y$$
$$5$$
$$3$$
$$7$$
The feasible region for the given lines is as indicated in the figure.
The values of $$z$$ at the corner points are:
At $$(0,0)$$, $$z = 0$$
At $$(0,5)$$, $$z = 5$$
At $$(3,7)$$, $$z = 3+7 = 10$$
At $$(6,4)$$, $$z = 6+4 = 10$$
At $$(6,0)$$, $$z = 6$$
Hence, the objective function $$z$$, has maximum value at every point of the segment joining two points of the feasible region.
The objective function $$z = 4x_{1} + 5x_{2}$$, subject to $$2x_{1} + x_{2}\geq 7, 2x_{1} + 3x_{2} \leq 15, x_{2}\leq 3, x_{1}, x_{2} \geq 0$$ has minimum value at the point.
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On x-axis
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On y-axis
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At the origin
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On the parallel to x-axis
Explanation
Value of $$z = 4x_{1} + 5x_{2}$$
Convert the given inequalities into equalities to get the corner points
$$2x_{1} + x_{2} = 7$$ ....... $$(i)$$
At $$x_1=0, x_{2}=7$$ and $$x_2=0, x_{1}=3.5$$
So, the corner points of $$(i)$$ are $$(0,7)$$ and $$(3.5,0)$$
$$2x_{1} + 3x_{2} = 15$$ ...... $$(ii)$$
At $$x_1=0, x_{2}=5$$ and $$x_{2}=0, x_{1}=7.5$$
So, the corner points of $$(i)$$ are $$(0,5)$$ and $$(7.5,0)$$
$$x_{2}=3$$ ...... $$(iii)$$
Plot these corner points on the graph paper and the line given in $$(iii)$$
The shaded part shows the feasible region.
At $$x_{2}=3, x_1=2$$ in $$(i)$$ and $$x_{1}=3$$ in $$(ii)$$
The corner points of the feasible region are $$(3.5,0), (7.5,0), (3,3)$$ and $$(2,3)$$
Corner points $$Z=4x_{2}+5x_{2}$$
$$(3.5,0)$$ $$14$$
$$(7.5,0)$$ $$30$$
$$(3,3)$$ $$27$$
$$(2,3)$$ $$23$$
Minimum value of $$Z= 14$$ and it lies on $$x$$-axis.
The objective function of LPP defined over the convex set attains its optimum value at
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atleast two of the corner points
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all the corner points
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atleast one of the corner points
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none of the corner points
Explanation
Let $$Z=ax+by$$ be the objective function
When $$Z$$ has optimum value(maximum or minimum), where the variables
$$x$$ and $$y$$ are subject to constraints described by linear inequalities, this optimum value must occur at a corner points of the feasible region.
Thus, the function attains its optimum value at one of the corner points.
Hence, C is correct.
The constraints
$$-{ x }_{ 1 }+{ x }_{ 2 }\le 1$$
$$-{ x }_{ 1 }+3{ x }_{ 2 }\le 9$$
$${x}_{1},{x}_{2}\ge 0$$ defines on
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Bounded feasible space
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Unbounded feasible space
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Both bounded and unbounded feasible
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None of the above
Explanation
Draw the following curves and find common area, which will give you feasible solutions.
So, our constraints are:
$$-{ x }_{ 1 }+{ x }_{ 2 }\le 1$$ -------------(i)
$$-{ x }_{ 1 }+3{ x }_{ 2 }\le 9$$ ----------(ii)
$${ x }_{ 1 }{ x }_{ 2 }\ge 0$$ -------------(iii)
We can see from the graph clearly that the constraints have unbounded feasible space
An article manufactured by a company consists of two parts $$X$$ and $$Y$$. In the process of manufacture of the part $$X$$. $$9$$ out of $$100$$ parts may be defective. Similarly $$5$$ out of $$100$$ are likely to be defective in part $$Y$$. Calculate the probability that the assembled product will not be defective.
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$$0.86$$
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$$0.864$$
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$$0.8456$$
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$$0.8645$$
Explanation
Let $$A=$$ Part $$X$$ is not defective
Probability of $$A$$ is $$P(A)=\dfrac {91}{100}$$
$$B=$$Part $$Y$$ is not defective.
Probability of $$B$$ is $$P(B)=\dfrac {95}{100}$$
Required probability $$=P(A\cap B)=P(A)P(B)=\dfrac {91}{100}\times \dfrac {95}{100}=\dfrac{8645}{10000}$$
The corner points of the feasible region determined by the system of linear constraints are $$(0, 10), (5, 5), (15, 15), (0, 20)$$. Let $$z=px+qy$$ where $$p, q > 0$$. Condition on p and q so that the maximum of z occurs at both the points $$(15, 15)$$ and $$(0, 20)$$ is __________.
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$$q=2p$$
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$$p=2p$$
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$$p=q$$
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$$q=3p$$
Explanation
Let $$z_{0}$$ be the maximum value of $$z$$ in the feasible region.
Since maximum occurs at both $$(15,15)$$ and $$(0,20)$$$, the value $$z_{0}$$ is attained at both $$(15,15)$$ and $$(0,20)$$.
$$\implies z_{0}=p(15)+q(15)$$ and $$ z_{0}=p(0)+q(20)$$
$$\implies p(15)+q(15)=p(0)+q(20)$$
$$\implies 15p=5q$$
$$\implies 3p=q$$
Hence, the answer is option (D).
Corner points of the bounded feasible region for an LP problem are $$A(0,5) B(0,3) C(1,0) D(6,0)$$. Let $$z = -50x + 20y$$ be the objective function. Minimum value of z occurs at ______ center point.
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$$(0,5)$$
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$$(1,0)$$
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$$(6,0)$$
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$$(0,3)$$
Explanation
We check the value of the $$z$$ at each of the corner points.
At $$A(0,5)-$$
$$z=-50x+20y=-50(0)+20(5)=100$$
At $$B(0,3)-$$
$$z=-50x+20y=-50(0)+20(3)=60$$
At $$C(1,0)-$$
$$z=-50x+20y=-50(1)+20(0)=-50$$
At $$D(6,0)-$$
$$z=-50x+20y=-50(6)+20(0)=-300$$
Hence, we see that $$z$$ is minimum at $$D(6,0)$$ and minimum value is $$-300$$.
The corner points of the feasible region are $$A(0,0),B(16,0),C(8,16)$$ and $$D(0,24)$$. The minimum value of the objective function $$z=300x+190y$$ is _______
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$$5440$$
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$$4800$$
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$$4560$$
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$$0$$
Explanation
We know that, for a cartesian polygon , the maximum value occurs at the corner points or vertices of the polygon.
Given $$z=300x+190y$$
By substituting $$A(0,0)$$ in the equation we get $$z=0$$
By substituting $$B(16,0)$$ in the equation we get $$z=4800$$
By substituting $$C(8,16)$$ in the equation we get $$z=5440$$
By substituting $$D(0,24)$$ in the equation we get $$z=4560$$
Hence the minimum value of Z occured at $$C(0,0)$$ with $$z=0$$
The given table shows the number of cars manufactured in four different colours on a particular day. Study it carefully and answer the question.
Colour
Number of cars manufactured
Vento
Creta
WagonR
Red
65
88
93
White
54
42
80
Black
66
52
88
Sliver
37
49
74
Which car was twice the number of silver Vento?
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Silver WagonR
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Red WagonR
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Red Vento
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White Creta
Explanation
The number of silver Vento car $$=37$$ (from the table)
Twice the number of silver Vento cars$$ = 2 \times 37=74$$
Now from table we can see that silver WagonR is only car type having $$74 $$ cars
A firm manufactures three products $$A,B$$ and $$C$$. Time to manufacture product $$A$$ is twice that for $$B$$ and thrice that for $$C$$ and if the entire labour is engaged in making product $$A,1600$$ units of this product can be produced.These products are to be produced in the ratio $$3:4:5.$$ There is demand for at least $$300,250$$ and $$200$$ units of products $$A,B$$ and $$C$$ and the profit earned per unit is Rs.$$90,$$ Rs$$40$$ and Rs.$$30$$ respectively.
Raw
material
Requirement per unit product(Kg)
A
Requirement per unit product(Kg)
B
Requirement per unit product(Kg)
C
Total availability (kg)
$$P$$
$$6$$
$$5$$
$$2$$
$$5,000$$
$$Q$$
$$4$$
$$7$$
$$3$$
$$6,000$$
Formulate the problem as a linear programming problem and find all the constraints for the above product mix problem.
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$$3{x}_{1}-4{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$
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$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$
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$$4{x}_{1}-3{x}_{2}=0$$ and $$4{x}_{2}-5{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$
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$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\le0$$
Explanation
Formulation of L.P Model
Let $${x}_{1},{x}_{2}$$ and $${x}_{3}$$ denote the number of units of products $$A,B$$ and $$C$$ to be manufactured .
Objective is to maximize the profit.
i.e., maximize $$Z=90{x}_{1}+40{x}_{2}+30{x}_{3}$$
Constraints can be formulated as follows:
For raw material $$P, 6{x}_{1}+5{x}_{2}+2{x}_{3}\le5,000$$
For raw material $$Q, 4{x}_{1}+7{x}_{2}+3{x}_{3}\le6,000$$
Product $$B$$ requires $$\frac{1}{2}$$ and product $$C$$ requires $${\left(\frac{1}{3}\right)}^{rd}$$ the time required for product $$A.$$
Then $$\frac{t}{2}$$ and $$\frac{t}{3}$$ are the times in hours to produce $$B$$ and $$C$$ and since $$1,600$$ units of $$A$$ will need time $$1,600t$$ hours, we get the constraint,
$$t{x}_{1}+\frac{t}{2}{x}_{2}+\frac{t}{3}{x}_{3}\le 1,600t$$ or
$${x}_{1}+\frac{{x}_{2}}{2}+\frac{{x}_{3}}{3}\le1,600$$ or
$$6{x}_{1}+3{x}_{2}+2{x}_{3}\le9,600$$
Market demand requires
$${x}_{1}\ge300, {x}_{2}\ge250,$$ and $${x}_{3}\ge200$$
Finally, since products $$A,B$$ and $$C$$ are to be produced in the ratio $$3:4:5,$$
$${x}_{1}:{x}_{2}:{x}_{3}::3:4:5$$
or $$\frac{{x}_{1}}{3}=\frac{{x}_{2}}{4},$$
and $$\frac{{x}_{2}}{4}=\frac{{x}_{3}}{5}.$$
Thus, there are two additional constraints
$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$
Equation of normal drawn to the graph of the function defined as $$f(x)=\dfrac{\sin x^2}{x}$$, $$x\neq 0$$ and $$f(0)=0$$ at the origin is?
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$$x+y=0$$
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$$x-y=0$$
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$$y=0$$
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$$x=0$$
Explanation
let slope of tangent=T, slope of normal=N
$$ T=\cfrac { dy }{ dx } =\cfrac { x.cos{ x }^{ 2 }\times 2x-sin{ x }^{ 2 } }{ { x }^{ 2 } } $$
$$ T\quad at(x=0)=\cfrac { cos{ x }^{ 2 }\times 2{ x }^{ 2 }-sin{ x }^{ 2 } }{ { x }^{ 2 } } $$
$$ x=0\quad \implies\quad 2cos{ x }^{ 2 }-\cfrac { sin{ x }^{ 2 } }{ { x }^{ 2 } } $$
$$ \quad \quad \quad (\cfrac { sinx }{ x } \quad at\quad x=1)$$
$$ \quad \implies 2-1=1\quad $$
$$ \quad \quad N\times T=-1$$
$$ \quad \quad N=-1$$
$$ at\quad x=0,y=0$$
$$ \quad y=mx$$
$$ \quad y=-x$$
$$ \quad x+y=0$$
The taxi fare in a city is as follows. For the first km the fare is $$Rs.10$$ and subsequent distance is $$Rs.6 / km.$$ Taking the distance covered as $$x \ km$$ and fare as $$Rs\ y$$ ,write a linear equation.
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0%
$$y=4+6x$$
0%
$$y=4+5x$$
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$$y=3+6x$$
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$$y=3+5x$$
Explanation
First $$km$$ fare =$$Rs.10$$
Subsequent distance fare=$$Rs\ 6/km$$
Then fare $$x\ km$$ of distance
$$y=(x-1) \times 6+10$$
$$y=6x-6+10$$
$$y=6x+4$$
$$A$$ is correct
The feasible region for anLPP is shown shaded in the figure. Find the maximum value of the objective function $$z=11x+7y$$.
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0%
35
0%
47
0%
21
0%
14
Explanation
the cover point of feasible
$$(0,3)$$, $$(0,5)$$, $$(3,2)$$
$$\rightarrow$$
$$Z(0,3)=21$$
$$Z(0,5)=35$$
$$Z(3,2)=47$$ = max value of $$Z$$
" the relation S is defined on $$N\times N\left ( a,b \right )S\left ( c,d \right )\Leftrightarrow bc\left ( a+d \right )=ad$$ in an equivalance " that statement is ?
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True
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False
The inequalities $$f(-1)\le -4,f(1) \le 0$$ & $$f(3) \le 5$$ are known to hold for $$f(x)=ax^{2}+bx+c$$ then the least value of $$'a'$$ is:
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$$-1/4$$
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$$-1/3$$
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$$1/4$$
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$$1/8$$
The problem associated with $$ LPP$$ is
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single objective function
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Double objective function
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No any objective function
0%
None
Explanation
The problem associated with LLP is single objective.
Shade region is represented by
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$$2x+5y \ge 80,x+y \le 20, x \ge 0,y \le 0$$
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$$2x+5y \ge 80,x+y \ge 20, x \ge 0,y \ge 0$$
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$$2x+5y \le 80,x+y \le 20, x \ge 0,y \ge 0$$
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$$2x+5y \le 80,x+y \le 20, x \ge 0,y \le 0$$
The maximum value of $$P=3x+4y$$ subject to the constraints $$x +y \le 40,x+2y \le 60,x \ge 0$$ and $$y \ge 0$$ is
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0%
$$120$$
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$$140$$
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$$100$$
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$$160$$
The maximum value of $$4x+5y$$ subject to the constraints $$x+y \le 20,x+2y \le 35,x-3y \le 12$$ is
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0%
$$84$$
0%
$$95$$
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$$100$$
0%
$$96$$
If $$x$$ is any real number, then which of the following is correct?
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$$\dfrac { x ^ { 2 } } { 1 + x ^ { 4 } } \geq \dfrac { 1 } { 2 }$$
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$$\dfrac { x ^ { 2 } } { 1 + x ^ { 4 } } \geq \dfrac { 1 } { 4 }$$
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$$\dfrac { x ^ { 2 } } { 1 + x ^ { 4 } } \leq \dfrac { 1 } { 2 }$$
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none of these
Explanation
We know that $$(x-\dfrac{1}{x})^{2}\geq 0$$
$$x^{2}+\dfrac{1}{x^{2}}-2\geq 0$$
$$x^{2}+\dfrac{1}{x^{2}}\geq 2$$
$$0 < \dfrac{1}{x^{2}+\dfrac{1}{x^{2}}}\leq \dfrac{1}{2}$$
$$0 \leq \dfrac{x^{2}}{x^{4}+1}\leq \dfrac{1}{2}$$ ($$\because x^{2}$$ and $$x^{4}$$ are positive or zero)
$$\dfrac{x^{2}}{x^{4}+1}\in [0,\dfrac{1}{2}]$$
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