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,x≥3,x+y≥5,x+2y≥6,and y≥0,is 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.

Corner point Z=-x+2y (6,0) Z=-6 B(4,1) z=-2 C(3,2) Z=1

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.