Question -
Answer -
On each day, let the factory manufacture x screws of type A andy screws of type B.
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as given below
| Screw A | Screw B | Availability |
| Automatic Machine (min) | 4 | 6 | 4 × 60 = 240 |
| Hand Operated Machine (min) | 6 | 3 | 4 × 60 = 240 |
The profit on a package of screws A is Rs 7 and on the packagescrews B is Rs 10
Hence, the constraints are
4x + 6y ≤ 240
6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem can be writtenas
Maximize Z = 7x + 10y …………. (i)
Subject to the constraints,
4x + 6y ≤ 240 …………. (ii)
6x + 3y ≤ 240 …………. (iii)
x, y ≥ 0 ……………… (iv)
The feasible region determined by the system of constraints isgiven below
A (40, 0), B (30, 20) and C (0, 40) are the corner points
The value of Z at these corner points are given below
| Corner point | Z = 7x + 10y | |
| A (40, 0) | 280 | |
| B (30, 20) | 410 | Maximum |
| C (0, 40) | 400 | |
The maximum value of Z is 410 at (30, 20)
Hence, the factory should produce 30 packages of screws A and 20packages of screws B to get the maximum profit of Rs 410