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Question
Refer to question 15. Determine the maximum distance that the man can travel.
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Solution
Referring to the solution of Question No.15, we have
Maximise Z = x + y subject to the constraints
Let 2x + 3y = 120
| x | 0 | 60 |
| y | 40 | 0 |
Let 8x + 5y = 400
| x | 0 | 50 |
| y | 80 | 0 |
2x + 3y ≤ 120 ......(i)
8x + 5y ≤ 400 ......(ii)
x ≥ 0, y ≥ 0
On solving eq. (i) and (ii) we get
x = `300/7` and y = `80/7`
Here, OABC is the feasible region whose corner points are O(0, 0), A(50, 0), `"B"(300/7, 80/7)` and C(0, 40).
Let us evaluate the value of Z
| Corner points | Value of Z = x + y | |
| O(0, 0) | Z = 0 + 0 = 0 | |
| A(50, 0) | Z = 50 + 0 = 50 km | |
| `"B"(300/7, 80/7)` |
Z = `300/7 + 80/7` = `380/7` = 54.3 km |
← Maximum |
| C(0, 40) | Z = 0 + 40 = 40 km |
Hence, the maximum distance that the man can travel is `54 2/7` km at `(300/7, 80/7)`.
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