Question

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

Types of Toys	Machines
	I	II	III
A	12	18	6
B	6	0	9

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Answer 1

Let the manufacturer makes x toys of type A and y toys of type B. We construct the following table:

Type of toys	Number of toys	Time of machine I (in min)	Time of machine II (in min)	Time of machine III (in min)	Profit (in Rs.)
A	x	12x	18x	6x	7.50x
B	y	6y	0y	9y	5y
Total	x + y	12x + 6y	18x + 0y	6x + 9y	7.50x + 5y
Maximum requires		6 x 60 = 360	6 x 60 = 360	6 x 60 = 360

Our problem is to maximize Z = 7.50x + 5y …(i) 

Subject to constraints 12x + 6y ≤ 360 ⇒ 2x + y ≤ 60 …(ii) 

18x ≤ 360 ⇒ x ≤ 20 …(iii) 

6x + 9y ≤ 360 ⇒ 2x + 3y ≤ 120 …(iv) 

x ≥ 0, y ≥ 0 …(v) 

Firstly, draw the graph of the line 2x + y = 60 

Secondly, draw the graph of the line 2x + 3y = 120 

Thirdly, draw the graph of the line x = 20 

On solving equations 2x + y = 60 and 2x + 3y = 120, we get C(15, 30) 

Similarly, solving the equations x = 20 and 2x + y = 60, we get B(20, 20). 

∴ Feasible region is OABCDO.

The corner points of the feasible region are A(20, 0), B(20, 20), C(15, 30) and D(0, 40). The values of Z at these points are as follows:

Thus, the maximum value of Z is 712.5 C(15, 30). 

Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.