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in Mathematics by (52.9k points)

A factory makes tennis rackets and cricket bats. A tennis racket takes 1x5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. If the profit on a racket and on a bat is Rs.20 and Rs.10 respectively, find the number of tennis rackets and crickets bats that the factory must manufacture to earn the maximum profit. Make it as an L.P.P. and solve graphically.

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by (266k points)
 
Best answer

Let the number of tennis rackets and cricket bats manufactured by factory be x and y respectively.
Here, profit is the objective function z.
z = 20x + 10y …(i)
We have to maximise z subject to the constraints

1 ×5x + 3y  42 …(ii) [Constraint for machine hour]
3x + y  24 …(iii) [Constraint for Craft man’s hour]

x≥0

y≥0

[Non-negative constraint]

Graph x = 0 and y = 0

calculating the value of x and y

Graph of 3x + y ≤  24 is the part of Ist quadrant in which origin lie
Hence, shaded area OACB is the feasible region.

For coordinate of C equation 1 × 5x + 3y = 42 and 3x + y = 24 are solved as

Now value of objective function z at each corner of feasible region is

value of z at each corner

Therefore, maximum profit is Rs200, when factory makes 4 tennis rackets and 12 cricket bats.

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