A manufacturer makes two types of toys A and B. Three machine are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

The machines I, II and III are available for a maximum of 3 hours, 2 hours and 2 hours 30 minutes respectively. The profit on each toy of type A is Rs 50, and that of type B is Rs 60. Formulate the above problem as a L.P.P and solve it graphically to maximize profit.
Let the manufacturer make x and y quantity of toy A and toy B respectively.
Max P = 50x + 60y
Subject to
20x + 10y ≤ 180
⇒ 2x + y ≤ 18 … (1)
10x + 20y ≤ 120
⇒ x +2 y ≤ 12 … (2)
10x + 30y ≤ 150
⇒ x + 3y ≤ 15 … (3)
x, y ≥ 0
Corner points of constraints are (1), (2), (3) are –
For 2x + y ≤ 18
At x = 0, x = 9,
At y = 0, y = 18

For x + 2y ≤ 12
At x = 0, y = 6,
At y = 0, x = 12

For x + 3y ≤ 15
At x = 0, y = 5,
At y = 0, x = 15



Clearly, Max Profit is Rs 520, at x = 8 and y = 2.
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.


