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

Asked by Sakshi | 1 year ago |  121

##### Solution :-

 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

Answered by Aaryan | 1 year ago

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