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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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Maximise $Z=3x+4y$,subject to the constraints: $x+y\leq 1,x\geq 0,y\geq 0$.

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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 coordinates of the vertices (corner points) of the shaded region are $O(0,0),A(1,0)$ and $B(0,1)$
The values of the objective function at these points are as follows :
Step 2:
For the point $O(0,0)$ the value of the objective function is $z=3x+4y\Rightarrow 3\times 0+4\times 0=0$
For the point $A(1,0)$ the value of the objective function is $z=3x+4y\Rightarrow 3\times 1+4\times 0=3$
For the point $B(0,1)$ the value of the objective function is $z=3x+4y\Rightarrow 3\times 0+4\times 1=4$
Step 3:
It is clearly that $z$ is maximum at $B(0,1)$.
The maximum value is $4$
answered Aug 19, 2013 by sreemathi.v
 
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