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?
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,
5x + 8y 
200
10x + 8y 
240
x 
0, y 
0
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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