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A company manufactures two type 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 available for assembling. The profit is 50 paisa each for type A and 60 paisa 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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Let the company manufacture x souvenirs of Type A and y souvenirs of Type B.

Therefore, x ≥ 0, y ≥ 0

The given information can be compiled in a table as follows:

The profit on Type A souvenirs is 50 paisa and on Type B souvenirs is 60 paisa. Therefore, profit gained on x souvenirs of Type A and y

souvenirs of Type B is Rs 0.50x and Rs 0.60y respectively.

Total Profit, Z = 0.5x + 0.6y

The mathematical formulation of the given problem is,

Max Z = 0.5x + 0.6y Subject to constraints,

Region 5x + 8y ≤ 200: line 5x + 8y = 200 meets axes at A(40, 0), B(0, 25) respectively. Region containing origin represents the solution of the inequation 5x + 8y ≤ 200 as (0, 0) satisfies 5x + 8y ≤ 200.

Region 10x + 8y ≤ 240: line 10x + 8y = 240 meets axes at C(24, 0), D(0, 30) respectively. Region containing origin represents the solution of the inequation 10x + 8y ≤ 240 as (0, 0) satisfies 10x + 8y ≤ 240.

Region x, y ≤ 0: it represents first quadrant.

The corner points of the feasible region are O(0, 0), B(0, 25), E(8, 20), C(24, 0).

The values of Z at these corner points are as follows:

The maximum value of Z is attained at E(8, 20).

Thus, 8 souvenirs of Type A and 20 souvenirs of Type B should be produced each day to get the maximum profit of Rs 16.

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