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Question
Show that the minimum of Z occurs at more than two points.
Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.
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Solution
The system of constraints is:
x + 2y ≤ 120 ...(i)
x + y ≥ 60 ....(ii)
x - 2y ≥ 0 ....(iii)
and x, y ≥ 0 ....(iv)
Let l1 : x + 2y = 120
l2 : x + y = 60
l3 : x - 2y = 0
It is observed that the feasible region CADE is bounded.
The coordinates of C, A, D, E are (60, 0), (120, 0), (60, 30), (40, 20) respectively.
Thus, we use the Corner Point Method to determine the maximum and minimum values of Z.
We have : Z = 5x + l0y
| Corner Point | Corresponding values of Z |
| (60, 0) | 300 (Minimum) |
| (120, 0) | 600 |
| (60, 30) | 600 |
| (40, 20) | 400 |
Hence Zmin = 300 at (60, 0) and Zmax = 600 at all points on the line segment joining the points (120, 0) and (60, 30).
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