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Question
Show that the minimum of Z occurs at more than two points.
Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
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Solution
The system of constraints is:
x + 2y ≥ 100 ....(i)
2x - y ≤ 0 ....(ii)
2x + y ≤ 200 ....(iii)
and x, y ≥ 0 ....(iv)
Let l1 : x + 2y = 100
l2 : 2x - y = 0
l3 : 2x + y = 200
The shaded region in the figure is the feasible region determined by the system of constraints (i) to (iv).
It is observed that the feasible region ECDB is bounded.
Thus, we use the Corner Point Method to determine the maximum and minimum values of Z.
We have, Z = x + 2y
The co-ordinates of E, C, D and B are (20, 40) (on solving x + 2y = 100 and 2x - y = 0),
(50, 100) (on solving 2x + y = 200 and 2x - y = 0), (0, 200) and (0, 50) respectively.
| Corner Point | Corresponding values of Z |
| (20, 40) | 100 |
| (50, 100) | 250 |
| (0, 200) | 400 (Maximum) |
| (0, 50) | 100 |
Hence Zmax = 400 at (0, 200) and Zmin = 100 at all points on the line segment joining the points (0, 50) and (20, 40).
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