Solve the Linear Programming Problem graphically: Maximum Z=9x+3y subject to  

             $2x+3y\le 13$

              $3x+y\le 5$

             $x, y\ge 0$         

For given inequality first we will convert it into equations,  

$2\mathrm{x}+3\mathrm{y}=13,3\mathrm{x}+\mathrm{y}=5,\mathrm{x}=0\text{ and }\mathrm{y}=0$

Region represented by $2\mathrm{x}+3\mathrm{y}\le 13$:

The line $2x+3y=13$ meets the coordinate axes at

$\mathrm{A}\left(\frac{13}{2},0\right) \mathrm{and} \mathrm{B}\left(0,\frac{13}{3}\right)$ respectively. By joining these points we obtain the line $2\mathrm{x}+3\mathrm{y}=13$

Clearly (0,0) satisfies the inequation $2x+3y\le 13$ . So, the region containing the origin represents the solution set of the inequation $2x+3y\le 13$

Region represented by $3x+y\le 5$:- 

The line $3x+y=5$ meets the coordinate axes at $C\left(\frac{5}{3},0\right)$ and $\mathrm{D}\left(\mathrm{0,5}\right)$ respectviely. By joining these points we obtain the line $3x+y=5$

Clearly (0,0) satisfies the inequation $3x+y\le 5$ . So, the region containing the origin represents the solution set of the inequation $3x+y\le 5$

Region represented by $x\ge 0$ and $y\ge 0$ is the 1st quadrant.

The feasible region determined by the system of constraints, $2x+3y\le \mathrm{13,3}x+y\le 5,x\le 0$  and $y\le 0,$ are shown in the shaded portion of the graph.

Notice that the corner points if the feasible region are O(0, 0), C$\left(\frac{5}{3}, 0\right)$, E$(\frac{2}{7},\frac{29}{7})$ and $B\left(0,\frac{13}{3}\right)$

  Now, we will find the values of Z at these corner points, it follows

We can see that the maximum value of the objective function Z is 15 which is at point C$\left(\frac{5}{3}, 0\right)$ and E$(\frac{2}{7},\frac{29}{7})$.