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Question
Refer to question 13. Solve the linear programming problem and determine the maximum profit to the manufacturer
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Solution
As per the solution of Question No.13
We have
Let 3x + 2y = 3600
| x | 0 | 1200 |
| y | 1800 | 0 |
Let x + 4y = 1800
| x | 0 | 1800 |
| y | 450 | 0 |
Maximise Z = 100x + 170y
Subject to the constraints
3x + 2y ≤ 3600 ......(i)
x + 4y ≤ 1800 .......(ii)
x ≥ 0, y ≥ 0
On solving equation (i) and (ii) we get
x = 1080 and y = 180
OABC is the feasible region whose corner points are O(0, 0), A(1200, 0), B(1080, 180), C(0, 450).
Let us evaluate the value of Z.
| Corner points | Value of Z = 100x + 170y | |
| O(0, 0) | Z = 100(0) + 170(0) = 0 | |
| A(1200, 0) | Z = 100(1200) + 0 = 120000 | |
| B(1080, 180) | Z = 100(1080) + 170(180) = 138600 |
← Maximum |
| C(0, 450) | Z = 170(450) = 76500 |
Hence, the maximum value of Z is 138600 at (1080, 180).
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