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 toy on the machines is given below:
Type of toy | Machines | ||
I | II | III | |
A | 12 | 18 | 6 |
B | 6 | 0 | 9 |
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.
Solution
Let x and y toys of type A and type B respectively be manufactured in a day.
The given problem can be formulated as follows.
Maximize z = 7.5x + 5y … (1)
subject to the constraints,
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (20, 0), B (20, 20), C (15, 30), and D (0, 40).
The values of z at these corner points are as follows.
Corner point | Z = 7.5x + 5y | |
A(20, 0) | 150 | |
B(20, 20) | 250 | |
C(15, 30) | 262.5 | → Maximum |
O(0, 40) | 200 |
The maximum value of z is 262.5 at (15, 30).
Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.