Step 1:

The objective function which is to maximized is given as

$Z=x+y$,subject to constraints $x+4y \leq 8,2x+3y\leq 12,3x+y\leq 9,x\geq 0,y\geq 0$

Now let us draw the line AB :$x+4y=8$,CD : $2x+3y=12$,EF :$3x+y=9$ on the graph.

Step 2:

Consider the line $x+4y=8$

Put $x=0,y=0$

$\Rightarrow 0\leq 8$

Which is true.

Hence the region $x+4y\leq 8$ lies below the line AB:$x+4y=8$

Consider the line$2x+3y=12$

Put $x=0,y=0$

$\Rightarrow 0\leq 12$

Which is true.

Hence the region $2x+3y\leq 12$ lies below the line CD :$2x+3y=12$

Consider the line$3x+y=9$

Put $x=0,y=0$

$\Rightarrow 0\leq 9$

Which is true.

Hence the region $3x+y\leq 9$ lies below the line $3x+y=9$

Step 3:

Clearly the origin $O$ is also included in the feasible region.

$\therefore$ The feasible region is $OEQA$,which is the shaded portion.

Step 4:

Let us obtain the maximum value of $Z=x+y$,subjected to the constraints as follows :

At the points $(x,y)$ the value of the objective function subjected to $Z=x+y$

At $O(0,0)$ the value of the objective function $Z=0+0=0$

At $A(0,2)$ the value of the objective function $Z=0+2=2$

At $Q(2.5,1.4)$ the value of the objective function $Z=2.5+1.4=3.9$

At $E(3,0)$ the value of the objective function $Z=3+0=3$

Clearly the maximum value is at $Q(2.5,1.4)$ and equal to 3.9