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Question

# A manufacturer produces nuts and bolts. It takes 1 hours of work on machine A and 3 hours on machine B to product 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 per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operates his machines for at the most 12 hours a day? From the above as a linear programming problem and solve it graphically.

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Solution

## Let x be the no. of nuts & y be the bolts.We have to maximize x & y if the factory worked at capacity of 12hours per day.Clearly, x≥y,y≥0Since we have only 12 hours of machine, we will use the following constraints:-x+3y≤123x+y≤12The profit of nuts is Rs.17.5 and bolts is Rs. 7 .We need to maximize 17.5x+7yWe can see that the feasible region is bounded and in the first quadrant.On solving the equations, x+3y=12 and 3x+y=12, we get x=3 and y=3.Therefore, feasible points are (0,0);(0,4);(3,3);(4,4)Values of z are(z=17.5x+7y)The maximum profit =73.5 which is when 3 packets each of nuts & bolts are produced.

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