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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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The feasible region for a LPP is shown in fig.12.9.Find the minimum value of $z=11x+7y$

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  • 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:
The given function is $z=11x+7y$
The lines subjected to the constraint are $x+y=5$ and $x+3y=9$
Let us first obtain the corner points of the feasible region.
Clearly the line $x+y=5$ touches the co-ordinate axes at $(5,0)$ and $(0,5)$.
The line $x+3y=9$ touches the coordinate axes at $(9,0)$ and $(0,3)$
Step 2:
The point of intersection of the two lines can be obtained by solving them.
$x+y=5$
$x+3y=9$
_______________
$-2y=-4$
$y=2$
$x+y=5$
$x+2=5$
$x=3$
The point of intersection is $(3,2)$
Step 3:
Now the coordinate of the corner points are $A(0,3),B(0,5)$ and $C(3,2)$
The values of objective function at these points are as follows :
For the point $A(0,3)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 0+7\times 3=21$
For the point $B(0,5)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 0+7\times 5=35$
For the point $C(2,3)$ the value of the objective function is $z=11x+7y\Rightarrow 11\times 3+7\times 2=47$
Hence the minimum value is $21$ at $(0,3)$
answered Aug 20, 2013 by sreemathi.v
edited Aug 20, 2013 by sreemathi.v
 

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