# Determine the maximum value of $z=11x+7y$ subject to the constraints: $2x+y\leq 6,x\leq 2,y\geq 0.$

$\begin{array}{1 1} 40 \\ 42 \\ 33 \\ 21.5 \end{array}$

Toolbox:
• Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Given :
$Z=11x+7y$
It is subject to constraints $2x+y \leq 6,x\leq 2,x\geq 0,y\geq 0$
Now let us convert the given inequalities into equation.
We obtain the following equation
$2x+y=6$
$x=2$
The region represented by $2x+y \leq 6$.The line $2x+y=6$ meets the coordinate axes at $A(3,0)$ and $B(0,6)$ respectively.Join these points to obtain the line $2x+y=6$
It is clear that $(0,0)$ satisfies the inequation $3x+5y \leq 15$.So the region containing the origin represents the solution set of the inequation $2x+y \leq 6$
Step 2:
The region represented by $x\leq 2$.The line is parallel to $y$-axis and meets the $x$-axis at $x=2$
It is clear $(0,0)$ satisfies this equation ,hence the region obtained contain the origin as its solution set.
Region represented by $x \geq 0$ and $y\geq 0$,Since every point in the first quadrant satisfies these inequations.
The coordinate of the vertices of the shaded region are $O(0,0),A(3,0),P(1,1.5)$ and $B(0,6)$
Step 3:
The values of the objective function at these points are given as follows:
For the point $O(0,0)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 0+7\times 0=0$
For the point $A(3,0)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 3+7\times 0=33$
For the point $P(1,1.5)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 1+7\times 1.5=21.5$
For the point $B(0,6)$ the value of the objective function is $z=11x+7y\Rightarrow 0+42=42$
Step 4:
It is clearly that $z$ is maximum at $B(0,6)$.
The maximum value is $42$