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Question
A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws 'A' at a profit of 70 P and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
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Solution
Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.
| Screw A | Screw B | Availability | |
| Automatic Machine (min) | 4 | 6 | 4 × 60 = 240 |
| Hand Operated Machine (min) | 6 | 3 | 4 × 60 = 240 |
\[4x + 6y \geq 240\]
The manufacturer can sell a package of screws 'A' at a profit of Rs 0.7 and screws 'B' at a profit of Re 1.
Total profit, Z = 0.7x + 1y
The mathematical formulation of the given problem is
Maximize Z = 0.7x + 1y
subject to the constraints,
x, y ≥ 0
First we will convert inequations into equations as follows:
4x + 6y = 240, 6x + 3y = 240, x = 0 and y = 0
Region represented by 4x + 6y ≤ 240:
The line 4x + 6y = 240 meets the coordinate axes at A1(60, 0) and B1(0, 40) respectively. By joining these points we obtain the line 4x + 6y = 240. Clearly (0,0) satisfies the 4x + 6y = 240. So,the region which contains the origin represents the solution set of the inequation 4x + 6y ≤ 240.
Region represented by 6x + 3y ≤ 240:
The line 6x + 3y = 240 meets the coordinate axes at C1(40, 0) and D1(0, 80) respectively. By joining these points we obtain the line 6x + 3y = 240. Clearly (0,0) satisfies the inequation 6x + 3y ≤ 240. So,the region which contains the origin represents the solution set of the inequation 6x + 3y ≤ 240.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 4x + 6y ≤ 240, 6x + 3y ≤ 240, x ≥ 0, and y ≥ 0 are as follows.
The corner points are C1(40, 0), E1(30, 20) and B1(0, 40).
The values of Z at these corner points are as follows.
| Corner point | Z = 7x + 10y |
| C1(40, 0) | 280 |
| E1(30, 20) | 410 |
| B1(0, 40) | 400 |
The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
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