CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 6 - MCQExams.com

If $$x+y \leq 2, x\leq 0, y\leq 0$$ the point at which maximum value of $$3x+2y$$ attained will be.
  • $$(0, 0)$$
  • $$\left ( \dfrac{1}{2}, \dfrac{1}{2} \right )$$
  • $$(0, 2)$$
  • $$(2, 0)$$
In figure 32, the shaded region within the triangle is the intersection of the sets of ordered pairs described by which of the following inequalities?
535617_3e2389f6159043e9941a24c688e7a582.png
  • y < x, x < 2
  • y < 2x, x < 2
  • y < 2x, x < 2, x > 0
  • y < 2x, y < 2, x > 0
  • y < 2x, x < 2, y > 0
The linear programming problem:
Maximize $$z={ x }_{ 1 }+{ x }_{ 2 }$$
Subject to constraints
$$\quad { x }_{ 1 }+2{ x }_{ 2 }\le 2000,{ x }_{ 1 }+{ x }_{ 2 }\le 1500,\quad { x }_{ 2 }\le 600,\quad { x }_{ 1 }\ge 0$$

  • No feasible solution
  • Unique optimal solution
  • A finite number of optimal solutions
  • Infinite number of optimal solutions
Solve the following LPP graphically. Maximize or minimize $$Z = 3x + 5y$$ subject to
$$3x - 4y \geq -12$$
$$2x - y + 2\geq 0$$
$$2x + 3y - 12\geq 0$$
$$0 \leq x \leq 4$$
$$y \geq 2$$.
  • Min. value $$19$$ at $$(5, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.
  • Min. value $$30$$ at $$(3, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.
  • Min. value $$19$$ at $$(3, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.
  • Min. value $$8$$ at $$(3, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.
Let $$P(-1, 0), Q(0, 0)$$ and $$R(3, 3\sqrt{3})$$ be three points. The equation of the bisector of the angle PQR is?
  • $$x+\sqrt{3}y=0$$
  • $$\sqrt{3}x+y=0$$
  • $$x+\dfrac{\sqrt{3}}{2}y=0$$
  • $$\dfrac{\sqrt{3}}{2}x+y=0$$
Use graph paper for this question:
  • Plot the points $$A(-4, 2)$$ and $$B(2, 4)$$
  • $${A}^{1}$$ is the image of $$A$$ when reflected in the line $$x=0$$. Write the co-ordinates of $${A}^{1}$$.
  • $${B}^{1}$$ is the image of $$B$$ when reflected in the line $$A{A}^{1}$$. Write the co-ordinates of $${B}^{1}$$.
  • Write the geometrical name of the figure $$AB{A}^{1}{B}^{1}$$
Minimise and Maximise $$Z = x + 2y$$ subject to the following constraints$$x + 2y \ge 100,  ~2x - y \le 0, ~y \le 200$$ and $$x, y \ge 0$$
  • Minimum $$200$$, Maximum $$400$$
  • Minimum $$100$$, Maximum $$500$$
  • Minimum $$400$$, Maximum $$500$$
  • Minimum $$100$$, Maximum $$400$$
Let $$X_1 are X_2$$ are optimal solution of a LPP, then 
  • $$x = \lambda x_1 + (1 - \lambda) x_2, \lambda \epsilon R$$ is also an optimum solution
  • $$X = \lambda x_1 + (1 - \lambda) X_2, 0 \le \lambda \le I$$ gives an option
  • $$ X = \lambda x_1 + (1 + \lambda). X_2, 0 \le \lambda \le 1$$ gives an optimal solution
  • $$ X = \lambda X_1 + (1 + \lambda) X_2, \lambda \epsilon R$$ gives an optimal
Which of the following statement id correct?
  • Every $$LLP$$ admits an optimal solution
  • An $$LLP$$ admits unique optimal solution
  • If an $$LPP$$ admits two optimal solutions it has infinite number of optimal solution
  • None of these
The point which provides the solution to the linear programming problem : Max P= 2x+3y subject to constraints :$$x\geq 0, y\geq 0,2x+2y\leq 9,2x+y\leq 7,x+2y\leq 8,$$ is
  • (3,2.5)
  • (2,3.5)
  • (2,2.5)
  • (1,3.5)
Feasible region is the set of points which satisfy
  • the objective function
  • all the given constraints
  • some of the given constraints
  • only one constraint
If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0), then the point of minimum z = 3x + 2y is 
  • (2, 2)
  • (0, 10)
  • (4, 0)
  • (3, 4)
The maximum value of z=10x + 6y subject to the constraints $$3x + y \leq 12, 2x + 5y \leq 34, x \geq 0, y \geq 0$$ is
  • 56
  • 65
  • 55
  • 66
The point of which the maximum value of x + y subject to the constraints $$x + 2y \leq 70, 2x + y \leq 95, x \geq 0, y \geq 0.$$
  • (30, 25)
  • (20,25)
  • (35, 20)
  • (40, 15)
The maximum value of z = 5x + 3y subject tot the constraints $$3x + 5y \leq 15, 5x + 2y \leq 10, x, y \geq 0$$ is 
  • 235
  • $$\dfrac{235}{9}$$
  • $$\dfrac{235}{19}$$
  • $$\dfrac{235}{3}$$
If the corner points of the feasible solution are (0, 0), (3, 0), (2, 1), $$\left ( 0, \dfrac{7}{3} \right )$$ the maximum value of z = 4x + 5y is
  • 12
  • 13
  • $$\dfrac{35}{3}$$
  • 0
0:0:1


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