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: Type 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 number of toys of type A be x the number of toys of type B be y.

Items	Number	Machine I	Machine II	Machine III	Profit
Type A	x	12 min	18 min	6 min	Rs. 7⋅5
Type B	y	6 min	0 min	9 min	Rs. 5
Max available time		6 hours = 360 min	6 hours = 360 min	6 hours = 360 min

The equation for machine A is given as,

12x+6y≤360 2x+y≤60

The equation for machine B is given as,

18x≤360 x≤20

The equation for machine C is given as.

6x+9y≤360 2x+3y≤120

We need to maximize the cost so we can use function which Maximize Z.

Maximize Z=7.5x+5y (1)

All constraints are given as,

2x+y≤60 x≤20 2x+3y≤120 x≥0,y≥0

2x+y≤60

x	30	0
y	0	60

2x+3y≤120

x	0	60
y	40	0

Plot the graph using equations of constraint.

Substitute the value of x and y in the equation (1) to check the max value.

Corner points	Value of Z
( 0,40 )	200
( 15,30 )	262.50
( 20,20 )	250
( 20,0 )	150

Thus, the cost will be maximum if the number of type A toys are 15 and the number of type B toys are 30 and maximum profit is 262.50.