Question

Refer to question 14. How many sweaters of each type should the company make in a day to get a maximum profit ? What is the maximum profit?

Answer 1

Refer to question 14, we have maximum 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).
$\begin{array}{cc}\text{Corner points}& \text{Corresponding value of Z=200x+120y}\\ (0,0)& 0\\ (200,0)& 40000\\ (150,150)& 150\times 200+120\times 150=48000\leftarrow Maximum\\ (100,200)& 100\times 200+120\times 200=44000\\ (0,100)& 120\times 100=12000\end{array}$
Hence, 150 sweaters of each type made by company and maximum profit =Rs 48000.