Question

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each purpose and the time (in minutes) required for each toy on the machines is given below:

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs. 7.50 and that on each toy of type B is Rs. 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Answer 1

Let the manufacturer produces `x` toys of type A and `y` toys of type B. Then

Maximise `Z=7.50x+5y`……………1
and constraints
`12x+6yge360implies2x+yle60`…………2
`18xle360impliesxle20`…………….3
`6x+9yle360implies2x+3yle120`.............4
`xge0,yge0`.................5
First, draw the graph of the line `2x+y=60`

Put `(0,0)` in the inequation `2x+yle60`
`2xx0+0le60implies0le60` (True)
Thus, the half plane contains the origin.
Now draw the graph of the line `2x+3y=120`

Put `(0,0)` in the inequation `2x+3yle120`,
`2xx0+3xx0le120`
`=0le120` (True)
Thus, the half plane contains the origin.
Now, draw the graph of theline `x=20`
Put `(0,0)` in the inequatioins `xle20,0le20` (True)
Thus, the half plane contains origin.
Since `x,yge0`. So the feasible region will be first quadrant.
The point of intersection of the equations `2x+y=60` and `2x+3y=120` is `C(15,30)`. Similarly, the pont of intersection of the equation `x=20` and `2x+y=60` is `B(20,20)`. Thus, the feasible region is OABCDO.
The vertices of feasible region are `A(20,0), B(20,20),C(15,30)` and `D(0,40)`. We find the value of Z at these vertices.

Thus, the maximum value of `Z` is Rs. 712.50 at point `C(15,30)`. Therefore to obtain the maximum cost Rs. 712.50, he makes 15 toys of type A and 30 toys of type B.