Question

A Linear Programming Problem is as follows:

Maximise / Minimise objective function

Z = 2x – y + 5

Subject to the constraints

3x + 4y ≤ 60

x + 3y ≤ 30

x ≥ 0, y ≥ 0

If the corner points of the feasible region are A(0, 10), B(12, 6), C(20, 0) and O(0, 0), then which of the following is true?

Options

• Maximum value of Z is 40

• Minimum value of Z is –5

• Difference of maximum and minimum values of Z is 35

• At two corner points, value of Z are equal

Answer 1

Minimum value of Z is –5

Explanation:

Corner Points	Value of Z = 2x – y + 5
A(0, 10)	Z = 2(0) – 10 + 5 = –5 (Minimum)
B(12, 6)	Z = 2(12) – 6 + 5 = 23
C(20, 0)	Z = 2(20) – 0 + 5 = 45 (Maximum)
O(0, 0)	Z = 0(0) – 0 + 5 = 5

So the minimum value of Z is –5.