    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.

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Solution

## 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. 7⋅5 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+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 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.  Suggest Corrections  1      Explore more