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 seater. The company can make atmost 300 sweaters and spend atmost 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.

Answer 1

Let the company manufactures x number of type A sweaters and y number of type B sweaters.
From the given information we see that cost to make a type A sweater is Rs 360 and cost to make a type B sweater is Rs 120.
Also, the company spend atmost Rs 72000 a day.
$\therefore 360x+120y\le 72000$
$\Rightarrow 3x+y\le \mathrm{600........}\left(i\right)$
Also, company can make atmost 300 sweaters.
$\therefore x+y\le \mathrm{300........}\left(ii\right)$
Further, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100 i.e.,
$x+100\ge y$
$\Rightarrow x-y\ge -\mathrm{100.....}\left(iii\right)$
Also, we have non-negative constraints for x and y i.e., $x\ge 0,y\ge \mathrm{0....}\left(iv\right)$
Hence, the company makes a profit of Rs 200 for each sweter of type A and Rs 120 for each sweater of type B i.e.,
Profit (Z)=200x+120y
Thus, the required LPP to maximise the profit is
Maximise Z =200x+120y is subject to constraints.
$3x+y\le 600x+y\le 300x-y\ge -100x\ge 0,y\ge 0$