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.

Max $Z = 200x + 120y$ subject to $x + y \leq 300, 3x + y \leq 600, y - x \leq 100, x, y \geq 0$

The correct answer is Option (2) → Max $Z = 200x + 120y$ subject to $x + y \leq 300, 3x + y \leq 600, y - x \leq 100, x, y \geq 0$

Let x and y be the number of sweaters of type A and type B respectively.

From the given information, we have the following constraints.

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

$x + y ≤ 300$ ...(ii);

$x+100≥y⇒y≤ x + 100$  ...(iii)

Profit $Z=200x + 120y$

Hence, the required LPP to maximise the profit is

Maximise $Z = 200x + 120y$

subject to the constraints

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