$A$ 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 maximize the profit to the company.

(A) Let the company manufacture $x$ number of type $A$ sweaters and $y$ number of type $B$ sweaters.
The company spends at most $Rs. 72000$ a day. Therefore,$360x + 120y \leq 72000$.
Dividing by $120$,we get $3x + y \leq 600 \dots (i)$.
The company can make at most $300$ sweaters in total. Therefore,$x + y \leq 300 \dots (ii)$.
The number of sweaters of type $B$ cannot exceed the number of sweaters of type $A$ by more than $100$. This means $y - x \leq 100$,which can be written as $-x + y \leq 100 \dots (iii)$.
The company makes a profit of $Rs. 200$ for each type $A$ sweater and $Rs. 120$ for each type $B$ sweater. The objective function to maximize profit is $Z = 200x + 120y$.
Thus,the $LPP$ formulation is:
Maximize $Z = 200x + 120y$
Subject to constraints:
$3x + y \leq 600$
$x + y \leq 300$
$-x + y \leq 100$
$x \geq 0, y \geq 0$