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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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 $42$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...