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Question
Solve the following Linear Programming Problem graphically:
Maximize: z = – x + 2y,
Subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
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Solution
The feasible region determined by the constraints, x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0 is given below.
Here, it be seen the can that feasible region is unbounded.
The values of Z at corner points A (3, 2), B (4, 1) and C (6, 0) are given below.
| Corner point | Corresponding value of Z = – x + 2y |
| A (3, 2) | 1 ( may or may not be the maximum value) |
| B (4, 1) | –2 |
| C (6, 0) | –6 |
Since the feasible region is unbounded, Z = 1 may or may not be the maximum value.
Now, we draw the graph of the inequality, – x + 2y > 1, and we check whether the resulting open half-plane has any point/s, in common with the feasible region or not.
Here, the resulting open half plane has points in common with the feasible region.
Hence, Z = 1 is not the maximum value. We conclude, Z has no maximum value.
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