Maximize $Z = 5x + 3y$ subject to$3x + 5y \le 15,5x + 2y \le 10,x \ge 0,y0.$

.: The system of constraints is :
$3x + 5y \le 15$ ...(1)

$5x + 2y \le 10$ ...(2)

and $x \ge 0,y \ge 0$ ...(3)

Let ${l_1}:3x + 5y = 15$

${l_2}:5x + 2y = 10$

The shaded region in the adjoining figure is

the feasible region determined by the system of constraints (1) to (3).

It is observed that the feasible region OCEB is bounded.

Thus, we use Corner Point Method to determine the maximum value of Z.

We have :$Z = 5x + 3y$ ...(4)

The co-ordinates of $O,C,E$ and B are (0, 0), (2, 0), $\left( {\cfrac{{20}}{{19}},\cfrac{{45}}{{19}}} \right)$

(on solving$3x + 5y = 15,5x + 2y = 10$ ) and (0, 3) respectively.

Hence,${Z_{\max }} = \cfrac{{235}}{{19}}$ at $\left( {\cfrac{{20}}{{19}},\cfrac{{45}}{{19}}} \right)$