Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Linear Programming

Question:

Minimize $Z = 3x + 5y$

Subject to constraints: $x, y \geq 0, x + 3y – 3 \geq 0, x + y – 2 \geq 0$

Options:

3

7

9

10

Correct Answer:

7

Explanation:

The correct answer is Option (2) → 7

The feasible region determined by the system of constraints, $x + 3y \geq 3, x + y \geq 2$, and $x, y \geq 0$ is given below:

Here, the feasible region is unbounded.

The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2)

The values of Z at these corner points are given below:

Corner Points

Coordinates

Objective Function $Z = 3x + 5y$

A

(3, 0)

9

B

(3/2, 1/2)

7 (Min.)

C

(0, 2)

10

As we wish to minimize Z, we are going to draw graph of $Z = 3x + 5y < 7$ and check whether the resulting half plane has any common points with the feasibe region or not.

As the inequality, $Z – 3x + 5y < 7$ passes through a corner point B ( 3/2, 1/2) without interfering the feasible region.

That means, the corner point B (3/2, 1/2) minimizes Z and the minimum value of Z is 7. When $x = 3/2, y = 1/2$.