A factory makes tennis rackets and cricket bats.

(i) Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

$\therefore 1.5 x+3 y \leq 42$                                          (1)

The craftsman’s time is not available for more than 24 hours.

$\therefore 3 x+y \leq 24$                                                (2)

The factory is to work at full capacity. Therefore,

$1.5 x+3 y=42$

$3 x+y=24$

On solving these equations, we obtain

x = 4 and y = 12

Thus, 4 rackets and 12 bats must be made.

(i) The given information can be complied in a table as follows.

$\therefore 1.5 x+3 y \leq 42$

$3 x+y \leq 24$

$x, y \geq 0$

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

$\therefore Z=20 x+10 y$

$3 x+y \leq 24$

$x, y \geq 0$

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

$\therefore Z=20 x+10 y$

The mathematical formulation of the given problem is

Maximize $Z=20 x+10 y$                   ...(1)

subject to the constraints,

$1.5 x+3 y \leq 42 \ldots$ (2)

$3 x+y \leq 24 \ldots$ (3)

$x, y \geq 0 \ldots$ (4)

The feasible region determined by the system of constraints is as follows.

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 as follows.

Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.