Linear Programming — Class 12 Maths Solution

Referring to Solution 14, We have the following conditions as per the question, maximise $Z = 200x + 120y$ subject to $x + y \le 300,$ $3x + y \le 600,$ $x - y \ge - 100,$ $x \ge 0,$ $y \ge 0$

On solving $x + y = 300$ and $3x + y = 600$, we get
$x = 150,y = 150$

On solving $x - y = - 100$ and $x + y = 300$, we get
$x = 100,y = 200$

From the shaded feasible region it is clear that coordinates of corner points are (0,0) ,

(200,0), (150,150), (100,200) and (0,100).

Hence, 150 sweaters of each type made by company and maximum profit =Rs.48000.