# Linear Programming Class 12 Mathematics Exam Questions

Please refer to Linear Programming Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

## Class 12 MathematicsExam Questions Linear Programming

Class 12 Mathematics students should read and understand the important questions and answers provided below for Linear Programming which will help them to understand all important and difficult topics.

Question. The common region determined by all the linear constraints of a LPP is called the _______ region.
A) Optimal
(B) Feasible
(C) Unbounded
(D) Infeasible

B

Question. In a LPP, the linear function which has to be minimized or maximized is called a linear __________ function.
(A) Objective
(B) Feasible
(C) Unbounded
(D) None of These

A

Question. If the feasible region for a LPP is _________, then the optimal value of the objective function Z = ax + by may or may not exist.
(A) Bounded
(B) optimal
(C) Unbounded
(D) None

C

Question. Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Let F = 4x + 6y be the objective function. The Minimum value of F occurs at
(A) (0, 2) only
(B) (3, 0) only
(C) the mid-point of the line segment joining the points (0, 2) and (3, 0) only
(D) any point on the line segment joining the points (0, 2) and (3, 0).

D

Question. In a LPP, the linear inequalities or restrictions on the variables are called _________.
(A) Decision Variable
(B) Constraints
(C) Corner Points
(D) Restrictions

B

Question. The feasible region for a LPP is shown in Fig. 12.9. Find the minimum value of Z = 11x + 7y.

Answer. Clearly the feasible region is bounded.
Find all corner points:

The minimum Value of Z is 21 at point (0,3)

Question. Maximise Z = 4x + y, subject to the constraints:
x + y ≤ 50
3x + y ≤ 90
x ≥ 0, y ≥ 0
The shaded region is the feasible region determined by the system of constraints. 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.

Hence, maximum value of Z is 120 at the point (30, 0).

Question. Maximise Z = 3x + 2y, subject to x + 2y 10, 3x + y 15, x, y ≥ 0.
The feasible region determined by the constraints, x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, and y ≥ 0, is as follows.

The corner points of the feasible region are A (5, 0), B (4, 3), and C (0, 5). The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is 18 at the point (4, 3).

Question. Maximise Z = 3x + 4y, subject to the constraints 😡 + y ≤ 4, x ≥ 0, y ≥ 0.
The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows.

The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4). The values of Z at these points are as follows.

Therefore, the maximum value of Z is 16 at the point B (0, 4).

Question. Minimise and Maximise Z = 5x + 10y subject to constraints x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥0, y ≥ 0