Question

Maximise Z = − x + 2y, subject to the constraints x≥3,x+y≥5,x+2y≥6,y≥0.

Answer 1

The feasible region determined by the constraints, is x≥3,x+y≥5,x+2y≥6,y≥0. as follows.

It can be seen that the feasible region is unbounded.
The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows.

As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, −x + 2y > 1, and check whether the resulting half plane has points in common with the feasible region or not. The resulting feasible region has points in common with the feasible region. Therefore, Z = 1 is not the maximum value. Z has no maximum value.