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 | Profit |
| A | 10 | 5 | - |
| B | 6 | 4 | - |
| Firm’s capacity per week | 1000 | 600 |
| Gadget | Foundry | Machine-shop | Profit |
| A | 10 | 5 | Rs 30 |
| B | 6 | 4 | Rs 20 |
| Firm’s capacity per week | 1000 | 600 |
Now, let x and y be the required weekly production of gadgets A and B.
Given:
Profit on each gadget A is = Rs 30
So, profit on x gadget of type A = 30x
Profit on each gadget B is = Rs 20
So, profit on x gadget of type B = 20y
Let total profit be ‘Z’
So, Z = 30x + 20y
Case I:
Given:
Production of one gadget A requires 10hours per week for foundry.
Production of one gadget B requires 6hours per week for foundry.
So, x units of gadget A requires 10x hours per week and
y unit of gadget B requires 6y hours per week.
Maximum capacity of foundry per week is 1000 hours, so => 10x + 6y ≤ 1000
Case II:
Given:
Production of one gadget A requires 5hours per week of machine-shop.
Production of one unit of gadget B requires 4hours per week of machine-shop.
So, x units of gadget A requires 5x hours per week and
y unit of gadget B requires 4y hours per week.
Maximum capacity of machine-shop per week is 600 hours, so = 5x + 4y ≤ 600
Hence, the required mathematical formulation of linear programming is:
Maximize Z = 30x + 20y
Subject to
10x + 6y ≤ 1000
5x + 4y ≤ 600
Where, x, y ≥ 0
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