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Question
A company manufactures two types of cardigans: type A and type B. It costs ₹ 360 to make a type A cardigan and ₹ 120 to make a type B cardigan. The company can make at most 300 cardigans and spend at most ₹ 72000 a day. The number of cardigans of type B cannot exceed the number of cardigans of type A by more than 200. The company makes a profit of ₹ 100 for each cardigan of type A and ₹ 50 for every cardigan of type B.
Formulate this problem as a linear programming problem to maximize the profit to the company. Solve it graphically and find the maximum profit.
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Solution
Let number of cardigans of type A and type B be x and y respectively.
To maximize : Z = 100x + 50y in ₹.
Subject to constraints: x ≥ 0, y ≥ 0,
x + y ≤ 300,
360x + 120y ≤ 72000
⇒ 3x + y ≤ 600,
y ≤ x + 200
⇒ y - x ≤ 200
| Corner Points | Value of Z (in ₹) |
| O(0, 0) | 0 |
| A(200, 0) | 20000 |
| B(150, 150) | 22500← Max. value |
| C(50, 250) | 17500 |
| D(0, 200) | 10000 |
Hence no. of cardigans of type A = 150
and no. of cardigans of type B = 150.
Also, maximum profit is ₹22500.
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