91Ó°ÊÓ

To solve the inequalities we convert the inequalities into linear equations and find the solutions of the equations to obtain the graph.

For shading the region, we choose a point. If the point satisfies the inequalities then the shaded region is towards the point otherwise, the shaded region is away from the point.

Given inequalities are

$\begin{array}{c}xâ‰\yen 0\\ y≤0\\ y≤2x+4\\ 3x+y≤9\end{array}$

Their respective linear equations are $x=0,y=0y=2x+4,3x+y=9.$and$4x+y=16.$

The points which satisfy the equation $y=2x+4$are $\left(0,4\right)$and$\left(-2,0\right)$.

The points, which satisfy the equation $3x+y=9$are $\left(0,9\right)$and $\left(3,0\right).$

We choose$\left(0,0\right)$to get the shaded region. The point $\left(0,0\right)$ satisfies all inequalities $xâ‰\yen 0,y≤0,y≤2x+4,3x+y≤9$

Therefore, the graph for the inequalities is

The feasible region is$OABC$, where co-ordinates of$O,A,B,C$are$\left(0,0\right),\left(3,0\right),\left(1,6\right),\left(0,4\right)$respectively.

$f(x,y)=2x+y.$

At point $O(0,0)$

$f(x,y)=0.$

At point $A(3,0)$

$\begin{array}{c}f(x,y)=2\left(3\right)+0\\ =6.\end{array}$

At point $B(1,6)$

$\begin{array}{c}f(x,y)=2\left(1\right)+6\\ =2+6\\ =8\end{array}$

At point $C(0,4)$

$\begin{array}{c}f(x,y)=2\left(0\right)+4\\ =4.\end{array}$