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Question
Show that the minimum of Z occurs at more than two points.
Maximise Z = – x + 2y, Subject to the constraints:
x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
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Solution
The system of constraints is:
x ≥ 3 ...(i)
x + y ≥ 5 ...(ii)
x + 2y ≥ 6 ...(iii)
and y ≥ 0 ...(iv)
Let l1 : x = 3
l2 : x + y = 5
l3 : x + 2y = 6
l4 : y = 0
The shaded region in the figure is the feasible region determined by the system of constraints (i) to (iv).
The corner points are C(6, 0), E(4, 1) and F(3, 2).
Applying the Corner Point Method, we have
| Corner Point | Corresponding values of Z |
| (6, 0) | -6 |
| (4, 1) | -2 |
| (3, 2) | 1 |
It appears that Zmax = 1 at (3, 2).
But the feasible region is unbounded, therefore, we draw the graph of the inequality -x + 2y > 1.
Since the half-plane represented by - x + 2y > 1 has points common with the feasible region.
∴ Zmax ≠ 1.
Hence, Z has no maximum value.
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