Question -
Answer -
The feasible region determined by the constraints, 
is given below
Here, 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 given below
| Corner point | Z = – x + 2y |
| A (6, 0) | Z = – 6 |
| B (4, 1) | Z = – 2 |
| C (3, 2) | Z = 1 |
Since the feasible region is unbounded, hence, z = 1 may or maynot be the maximum value
For this purpose, we graph the inequality, – x + 2y > 1, andcheck whether the resulting half plane has points in common with the feasibleregion or not.
Here, the resulting feasible region has points in common withthe feasible region
Hence, z = 1 is not the maximum value.
Z has no maximum value.