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.

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.

Philoid · Accepted Answer

Let the company manufactures x number of type A sweaters and y number of type B. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=200x+120y

Now, we have to maximize the profit, i.e., maximize Z=200x+120y

The constraints so obtained, i.e., subject to the constraints,

The company spends at most Rs 72000 a day.
∴ 360x + 120y ≤ 72000

Divide throughout by 120, we get
=> 3x+y≤ 600 …(i)
Also, company can make at most 300 sweaters.
∴ x+y≤ 300 …(ii)
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100

i.e., y-x≤ 100

⇒ y≤ 100+x……….(iii)
And x≥0, y≥0 [non-negative constraint]

So, to maximize profit we have to maximize Z=200x+120y, subject to

3x+y≤ 600

x+y≤ 300

y≤ 100+x

x≥0, y≥0