Question

A manufacturer makes two types of toys A and B. Three machine are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
$\begin{array}{cccc}\text{Types of Toys}& & \text{Machines}& \\ & I& II& III\\ A& 20& 10& 10\\ B& 10& 20& 30\end{array}$
The machines I, II and III are available for a maximum of 3 hours, 2 hours and 2 hours 30 minutes respectively. The profit on each toy of type A is Rs 50 and that of type B is Rs 60. Formulate the above problem as a L.P.P. and solve it graphically to maximize profit.

Answer 1

Let the manufacturer makes x and y number of toy A and toy B, respectively.
To maximize: Z =Rs (50x +60y )
Subject to constraints : $x,y\ge 0;$
$20x+10y\le \mathrm{180...}\left(1\right),$
$10x+20y\le \mathrm{120...}\left(2\right),$
$10x+30y\le \mathrm{150...}\left(3\right)$
$\begin{array}{cc}\text{Corner Points}& \text{Volue of Z(in Rs)}\\ O(0,0)& 0\\ A(0,5)& 300\\ B(6,3)& 480\\ C(8,2)& 520\to \text{Max.}\\ D(9,0)& 450\end{array}$
Hence, the maximum profit is Rs 520.
And, nos. of toy A and B are respectively. 8 and, 2.