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.

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Let the number of toys of type A be x the number of toys of type B be y.

Items Number Machine IMachine IIMachine IIIProfit
Type A x 12 min 18 min 6 minRs. 75
Type B y 6 min 0 min 9 minRs. 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+6y360 2x+y60

The equation for machine B is given as,

18x360 x20

The equation for machine C is given as.

6x+9y360 2x+3y120

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

MaximizeZ=7.5x+5y (1)

All constraints are given as,

2x+y60 x20 2x+3y120 x0,y0


x 30 0
y 0 60


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 pointsValue 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.

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