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Question
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:
| Gadget | Foundry | Machine-shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Firm's capacity per week | 1000 | 600 |
The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
Solution
Let x and y number of gadgets A and B respectively being produced in order to maximize the profit.
Since, each unit of gadget A takes 10 hours to be produced by machine A and 6 hours to be produced by machine B and each unit of gadget B takes 5 hours to be produced by machine A and 4 hours to be produced by machine B.
Therefore, the total time taken by the Foundry to produce x units of gadget A and y units of gadget B is
Hence, 10x + 6y ≤ 1000.
This is our first constraint.
The total time taken by the machine-shop to produce x units of gadget A and y units of gadget B is 5x + 4y. This must be less than or equal to the total hours available.
Hence, 5x + 4y ≤ 600
This is our second constraint.
Since x and y are non negative integers, therefore
Let Z denotes the total cost
Therefore, Z= Rs (30x + 20y)
Hence, the above LPP can be stated mathematically as follows:
Maximize Z = 30x + 20y
subject to
10x + 6y ≤ 1000,
5x + 4y ≤ 600
x, y ≥ 0
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