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 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.

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.

$\therefore$ $360x + 120y \le 72000$
$\Rightarrow$ $3x + y \le 600$ …….(i)

Also, company can make atmost 300 sweaters
$\therefore$ $x + y \le 300$ …….(ii)

Further, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100 i.e.,

$x + 100 \ge y$

$\Rightarrow$ $x - y \ge - 100$ ……..(iii)

Also, We have the following conditions as per the question, non-negative constraints for $x$ and $y$ i.e.,

$x \ge 0,y \ge 0$ ……(iv)

Hence, the company makes a profit of Rs.200 for each sweater 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.

$3x + y \le 600$

$x + y \le 300$

$x - y \ge - 100$

$x \ge 0,y \ge 0$