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.