A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman’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 Rs20 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.

Let the number of rackets and the number of bats to be made be  x  and  y 

respectively.

The given information can be tabulated as below:

In a day, the machine time is not available for more than  42  hours.

$\therefore 1.5 \mathrm{x} + 3 \mathrm{y} ⩽ 42$

In a day, the craftsman's time can not be  more than  24  hours.

$\therefore 3 \mathrm{x} + \mathrm{y} ⩽ 24$

Let the total profit be Rs. Z.

The profit on a racket is Rs. 20  and on a bat is Rs. 10.

$\therefore \mathrm{Z} = 20 \mathrm{x} + 10 \mathrm{y}$

Thus, the given linear programming problem can be stated as follows:

Maximise   Z = 20 x + 10 y           ...........( i )

Subject to 

1.5 x + 3 y $⩽$ 42                         ...........( ii )

3 x + y $⩽$ 24                               ..........( iii )

x, y $⩾$ 0                                      ..........( iv )

The feasible region can be shaded in the graph as below:

The corner points are  A ( 8, 0 ),  B ( 4, 12 ),  C ( 0, 14 )   and   O ( 0, 0 ).

The values of  Z  at these corner points are tabulated as follows:

The maximum value of Z is 200,  which occurs at  x = 4  and  y = 12.

Thus, the factory must produce  4  tennis rackets  and 12 cricket bats to earn the maximum profit of Rs. 200.