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Question
Maximise Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0
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Solution
Given that: Z = 3x + 4y and the constraints x + y ≤ 1, x ≥ 0, y ≥ 0
Let x + y = 1
| x | 1 | 0 |
| y | 0 | 1 |
The shaded area OAB is the feasible region determined by x + y ≤ 1, x ≥ 0, y ≥ 0
The feasible region is bounded.
So, maximum value will occur at the corner points O(0, 0), A(1, 0), B(0, 1).
Now, evaluating the value of Z, we get
| Corner points | Value of Z | |
| O(0, 0) | 3(0) + 4(0) = 0 | |
| A(1, 0) | 3(1) + 4(0) = 3 | |
| B(0, 1) | 3(0) + 4(1) = 4 | ← Maximum |
Hence, the maximum value of Z is 4 at (0, 1).
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