Question

Maximise Z = x + y subject to x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0, y ≥ 0.

Answer 1

Given: Z = x + y subject to x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0, y ≥ 0.

Constructing a constrain table for the above, we have

Table for x + 4y = 8

On solving equations x + 4y £ 8 and 3x + y £ 9, we get

x = 28/11 and y = 15/11

Here, it’s seen that OABC is the feasible region whose corner points are O(0, 0), A(3, 0), B(28/11, 15/11) and C(0, 2).

Now, let’s evaluate the value of Z

From the above table it’s noticed that the maximum value of Z is 3.9

Therefore, the maximum value of Z is 3.9 at (28/11, 15/11).