Browse Questions

# Maximise the function $Z=11x+7y$,subject to the constraints : $x\leq 3,y\leq 2,x\geq 0,y\geq 0.$

$\begin{array}{1 1}33 \\ 14 \\ 47 \\ 0 \end{array}$

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 :$z=11x+7y$,subject to constraints $x\leq 3,y\leq 2,x\geq 0,y\geq 0$.
Now let us convert the inequalities to equation.
We obtain the following equation:
$x=3$ and $y=2$
Step 2:
The line $x=3$ is parallel to the $y$-axis and it meets the $x$-axis at $(3,0)$
The line $y=2$ is parallel to the $x$-axis and it meets the $y$-axis at $(0,2)$
The two lines meets at the point $(3,2)$.Hence the coordinates of the vertices (corner points) are $O(0,0),A(3,0),B(3,2),C(0,2)$
Step 3:
The values of the objective function at there points are 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 $B(3,2)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 3+7\times 2=47$
For the point $C(0,0)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 0+7\times 2=14$
Step 4:
Hence clearly $z$ has maximum value at $B(3,2)$ and the maximum value is 47.