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Question
Determine the maximum value of Z = 11x + 7y subject to the constraints : 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.
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Solution
Given that: Z = 11x + 7y and the constraints 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0
Let 2x + y = 6
| x | 0 | 3 |
| y | 6 | 0 |
The shaded area OABC is the feasible region determined by the constraints
2x + y ≤ 6
x ≤ 2
x ≥ 0
y ≥ 0
The feasible region is bounded.
So, maximum value will occur at a corner point of the feasible region.
Corner points are (0, 0), (2, 0), (2, 2) and (0, 6).
Now, evaluating the value of Z, we get
| Corner points | Value of Z | |
| O(0, 0) | 11(0) + 7(0) = 0 | |
| A(2, 0) | 11(2) + 7(0) = 22 | |
| B(2, 2) | 11(2) + 7(2) = 36 | |
| C(0, 6) | 11(0) + 7(6) = 42 | ← Maximum |
Hence, the maximum value of Z is 42 at (0, 6).
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