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Question

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs.5 each for type A and Rs.6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

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Solution


Let number of souvenirs A to be made is X and number of souvenirs B to be made is Y

Since, souvenirs A requires 5 minutes and souvenirs B requires 8 minutes for cutting. Also, there is maximum 3 hours 20 minutes (200 minutes) for cutting.
5X+8Y200 ...(1)

Since, souvenirs A requires 10 minutes and souvenirs B requires 8 minutes for assembling. Also, there is maximum 4 hours (240 minutes) for assembling.
10X+8Y240

5X+4Y120 ...(2)

Since, count of objects can't be negative.
X0,Y0 ...(3)

We have to maximize profit of the company.
Here, profit on souvenirs A is 5 Rs and on souvenirs B is 6 Rs

So, objective function is Z=5X+6Y

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points Value of Z=5X+6Y
A (0,25) 150
B (8,20) 160 (maximum)
C (24,0) 120
Hence, company should produce 8 souvenirs A and 20 souvenirs B in a day to maximize his profit. Also, maximum profit will be 160 Rs

815839_846994_ans_340e6ddf83e24dc08ac2b9a00fe0b4d0.png

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