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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)

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(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 $A(0,3)$

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