Question

Maximize : Z = 5x + 7x,
Subject to :x, + x, < 4
5x, + 8x, < 30
10x, + 7x, < 35
X, X, 20
<
.​

Answer 1

Answer:

z = 12 x

so maximum value is the upper limit of the interval in each case

Answer 2

Answer:

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0. Answer: ... Therefore, the maximum value of Z is 235/19 at the point (20/19, 45/19). ... Maximise Z = − x + 2y, subject to the constraints:.