Advertisement Remove all ads

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 - Mathematics

Graph
Sum

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.

Advertisement Remove all ads

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.

  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×