Question -
Answer -
Let the manufacturer produce x package of nuts and y package ofbolts. Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as given below
| Nuts | Bolts | Availability |
| Machine A (h) | 1 | 3 | 12 |
| Machine B (h) | 3 | 1 | 12 |
The profit on a package of nuts is Rs 17.50 and on a package ofbolts is Rs 7
Hence, the constraints are
x + 3y ≤ 12
3x + y ≤ 12
Then, total profit, Z = 17.5x + 7y
The mathematical formulation of the given problem can be writtenas follows
Maximize Z = 17.5x + 7y …………. (1)
Subject to the constraints,
x + 3y ≤ 12 …………. (2)
3x + y ≤ 12 ………… (3)
x, y ≥ 0 …………….. (4)
The feasible region determined by the system of constraints isgiven below
A (4, 0), B (3, 3) and C (0, 4) are the corner points
The values of Z at these corner points are given below
| Corner point | Z = 17.5x + 7y | |
| O (0, 0) | 0 | |
| A (4, 0) | 70 | |
| B (3, 3) | 73.5 | Maximum |
| C (0, 4) | 28 | |
Therefore, Rs 73.50 at (3, 3) is the maximum value of Z
Hence, 3 packages of nuts and 3 packages of bolts should beproduced each day to get the maximum profit of Rs 73.50