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Question 9:
Maximise Z = − x + 2y, subject to the constraints:
.
The feasible region determined by the constraints, 
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 |
A(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.
