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Question
Maximise Z = x + y subject to x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0, y ≥ 0.
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Solution
We are given that Z = x + y subject to the constraints
x + 4y ≤ 8 ......(i)
| x | 0 | 8 |
| y | 2 | 0 |
2x + 3y ≤ 12 ......(ii)
| x | 0 | 6 |
| y | 4 | 0 |
3x + y ≤ 9 ......(iii)
| x | 0 | 3 |
| y | 9 | 0 |
x ≥ 0, y ≥ 0
On solving equation (i) and (iii) we get
x = `28/11` and y = `15/11`
Here, OABC is the feasible region whose corner points are O(0, 0), A(3, 0), `"B"(28/11, 15/11)`, C(0, 2)
Let us evaluate the value of Z
| Corner points | Value of Z = x + y | |
| O(0, 0) | Z = 0 + 0 = 0 | |
| A(3, 0) | Z = 3 + 0 = 3 | |
| `"B"(28/11, 15/11)` | Z = `28/11 + 15/11 = 43/11` = 3.9 | ← Maximum |
| C(0, 2) | Z = 0 + 2 = 2 |
Hence, the maximum value of Z is 3.9 at `(28/11, 15/11)`.
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