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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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The feasible region for a LPP is shown in fig 12.10.Evaluate $Z=4x+y$ at each of the corner points of this region.Find the minimum value of $Z$,if it exists.

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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 function $Z=4x+y$ has constraints between the lines $x+2y=4$ and $x+y=3$
First let us obtain the corner points of the feasible region.
Clearly the line $x+2y=4$ touches the coordinate axes at $(4,0)$ and $(0,2)$
The line $x+y=3$ touches the coordinate axes at $(3,0)$ and $(0,3)$
Step 2:
The point of intersection can be obtained by solving these lines for $x$ and $y$
$x+2y=4$-----(1)
$x+y=3$------(2)
________________
(1)-(2)$\Rightarrow y=1$
$x+2y=4$
$x+2\times 1=4$
$x=4-2$
$x=2$
The point of intersection is $(2,1)$
Step 3:
The point of intersection is $(2,1)$
As shown in the figure the corner points of the feasible points are
$A(0,3),B(2,1)$ and $(4,0)$
The values of the objective function at these points are given as follows :
Step 4:
For the point $A(0,3)$ the value of the objective function is $z=4x+y\Rightarrow 4\times 0+ 3=3$
For the point $A(2,1)$ the value of the objective function is $z=4x+y\Rightarrow 4\times 2+ 1=9$
For the point $C(4,0)$ the value of the objective function is $z=4x+y\Rightarrow 4\times 4+ 0=16$
Hence the minimum value is 3 at the point $A(0,3)$
answered Aug 20, 2013 by sreemathi.v
 

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