How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

Question

A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

Answer 1

Let’s assume x and y to be the number of sweaters of type A and type B respectively.

From the question, the following constraints are:

360x + 120y ≤ 72000 ⇒ 3x + y ≤ 600 … (i)

x + y ≤ 300 … (ii)

x + 100 ≥ y ⇒ y ≤ x + 100 … (iii)

Profit: Z = 200x + 120y

Therefore, the required LPP to maximize the profit is

Maximize Z = 200x + 120y subject to constrains

3x + y ≤ 600, x + y ≤ 300, y ≤ x + 100, x ≥ 0, y ≥ 0.

Maximize Z = 200x + 120y subject to constrains

3x + y ≤ 600 …. (i)

x + y ≤ 300 …. (ii)

x – y ≤ -100 …. (iii)

x ≥ 0, y ≥ 0

Now, let’s construct a constrain table for the above

Table for (i)

Next, solving equation (i) and (iii) we get

x = 100 and y = 200

On solving equation (i) and (ii), we get

x = 150 and y = 150

It’s seen that the shaded region is the feasible region whose corner points are O(0, 0), A(200, 0), B(150, 150), D(0, 100).

Evaluating the value of Z, we have

From the above table it’s seen that the maximum value is 48000.

Therefore, the maximum value of Z is 48000 at (150, 150) which means 150 sweaters of each type.