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.

#### 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.