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Corner points of the feasible region determined by the system of linear constraints are $(0,3),(1,1)$ and $(3,0).$

Let Z=px+qy,where p,q>0.Condition on p and q so that the minimum of Z occurs at (3,0) and (1,1) is$(A)\;p=2q\quad(B)\;p=\frac{q}{2}\quad(C)\;p=3q\quad(D)\;p=q$

Can you answer this question?

Toolbox:
• 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 corner points are $(0,3),(1,1)$ and $(3,0)$
The objective function $Z=px+qy$
At the points $(x,y)$ the objective function subject to $Z=px+qy$
Step 2:
At $(0,3)$ the objective function $Z=px+qy\Rightarrow p\times 0+q\times 3=3q$
At $(1,1)$ the objective function $Z=px+qy\Rightarrow p\times 1+q\times 1=p+q$
At $(3,0)$ the objective function $Z=px+qy\Rightarrow p\times 3+q\times 0=3p$
Step 3:
It is given that the minimum of $Z$ occurs at $(3,0)$ and $(1,1)$
(i.e)$3p=p+q$
$\Rightarrow 2p=q$
$\Rightarrow p=\large\frac{q}{2}$
Hence $B$ is the correct option.
answered Aug 27, 2013