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\[(A)\;p=2q\quad(B)\;p=\frac{q}{2}\quad(C)\;p=3q\quad(D)\;p=q\]

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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 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.

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