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Question

# A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs.17.50 per package on nuts and Rs.7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?

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Solution

## Let number of package of nuts to be made is X and number of package of bolts to be made is YSince, nuts package requires 1 hours and bolts package requires 3 hours of machine A work time. Also, there is maximum 12 hours of machine A time available.∴X+3Y≤12 ...(1)Since, nuts package requires 3 hours and bolts package requires 1 hours of machine B work time. Also, there is maximum 12 hours of machine B time available.∴3X+Y≤12 ...(1)Since, count of package can't be negative.∴X≥0,Y≥0 ...(3)We have to maximize profit of the manufacturer.Here, profit on nuts package is 17.5 Rs and on bolts package is 7 RsSo, objective function is Z=17.5X+7YPlotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image. Corner points Value of Z=17.5X+7Y A (0,4)28 B (3,3) 73.5 (maximum) C (4,0) 70Hence, manufacturer should produce 3 packages of each to maximize his profit.

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