A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs 7 profit and B at a profit of Rs 4. Find the production level per day for maximum profit graphically.
Solution
Let the numbers of units of products A and B to be produced be x and y, respectively
| Product | Machine | |
| I (h) | II (h) | |
| A | 3 | 3 |
| B | 2 | 1 |
Total profit: Z = 7x + 4y
We have to maximise Z = 7x + 4y, which is subject to constraints.
3x+2y≤12 (Constraint on machine I)
3x+y ≤9 (Constraint on machine II)
⇒x≥0 and y≥0
The given information can be graphically expressed as follows:
Values of Z = 7x + 4y at the corner points are as follows:
| Corner Point | Z = 7x + 4y |
| (0, 6) | 24 |
| (2, 3) | 26 |
| (3, 0) | 21 |
Therefore, the manufacturer has to produce 2 units of product A and 3 units of product B for the maximum profit of Rs 26.
