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Question
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
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Solution
The shaded region is the feasible region determined by the constraints x ≤ 3, y ≤ 2, x ≥ 0, y ≥.
The feasible region is bounded with four corners O(0, 0), A(3, 0), B(3, 2) and C(0, 2).
So, the maximum value can occur at any corner
Let us evaluate the value of Z.
| Corner points | Value of Z | |
| O(0, 0) | 11(0) + 7(0) = 0 | |
| A(3, 0) | 11(3) + 7(0) = 33 | |
| B(3, 2) | 11(3) + 7(2) = 47 | ← Maximum |
| C(0, 2) | 11(0) + 7(2) = 14 |
Hence, the maximum value of the function Z is 47 at (3, 2).
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