Question

Find the maximum value of Z = 3x + 5y, subject to the constraints

– 2x + y ≤ 4, x + y ≥ 3, x – 2y ≤ 2, x ≥ 0 and y ≥ 0

Answer 1

It is given that

Z = 3x + 5y, subject to the constraints

– 2x + y ≤ 4, x + y ≥ 3, x – 2y ≤ 2, x ≥ 0 and y ≥ 0

Draw the line -2x + y = 4, x + y = 3 and x – 2y = 2 and shaded region which is satisfied by above inequalities

We know that the feasible region is bounded

A (8/3, 1/3), B (0, 3) and C (0, 4) are the corner points

Value of Z at A (8/3, 1/3)

Z = 3 × 8/3 + 5 × 1/3 = 29/3 = 9 2/3

Value of Z at B (0, 3)

Z = 3 × 0 + 5 × 3 = 15

Value of Z at C (0, 4)

Z = 3 × 0 + 5 × 4 = 20

Hence, the maximum value of Z is 9 2/3 which occurs at A (8/3, 1/3).