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Question
Refer to question 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit.
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Solution
Referring to the solution of Question No.14,
We have Maximise Z = 200x + 120y
Subject to the constraints
x + y ≤ 300 ......(i)
3x + y ≤ 600 ......(ii)
x – y ≥ – 100 ......(iii)
x ≥ 0, y ≥ 0
On solving equation (i) and (iii)
We have x = 100, y = 200
On solving eq. (i) and (ii)
We get x = 150, y = 150
Let x + y = 300
| x | 0 | 300 |
| y | 300 | 0 |
Let 3x + y = 600
| x | 0 | 200 |
| y | 600 | 0 |
Let x + y = –100
| x | 0 | –100 |
| y | 100 | 0 |
Here, the shaded region is the feasible region whose corner points are O(0, 0), A(200, 0), B(150, 150), C(100, 200), D(0, 100).
Let us evaluate the value of Z.
| Corner points | Value of Z = 200x + 120y | |
| O(0, 0) | Z = 200(0) + 120(0) = 0 | |
| A(200,0) | Z = 200(200) + 120(0) = 40000 | |
| B(150, 150) | Z = 200(150) + 120(150) = 48000 | ← Maximum |
| C(100, 200) | Z = 200(100) + 120(200) = 44000 | |
| D(0, 100) | Z = 200(0) + 120(100) = 12000 |
Hence, the maximum value of Z is 48000 at (150, 150)
i.e., 150 sweaters of each type.
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