Question

A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws 'A' while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell all the screws 'A' at a profit of 70 paise and screws 'B' at a profit of Rs1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit ? Formulate the above LPP and solve it graphically and find the maximum profit.

Answer 1

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

To maximize : Z = (0.70 x+ y) in Rs

Subject to constraints : x, y $\ge$ 0,

4x+6y $\le$ 240 $\Rightarrow$ 2x+3y $\le$ 120,

6x+3y $\le$ 240 $\Rightarrow$ 2x+ y $\le$ 80

$\begin{array}{cc}\text{Corner Points}& \text{Value of Z}\\ \text{A}(40,0)& 28\\ \text{B}(30,20)& 41\\ \text{C}(0,40)& 40\\ \text{O}(0,0)& 0\end{array}$

Hence maximum profit of Rs 41 is obtained at B (30,20). Therefore, number of packets of screw of type A is 30 and that of type B is 20.