Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Linear Programming
0 votes

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?

1 Answer

0 votes
  • 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 $(1,1)$
$\Rightarrow 2p=q$
$\Rightarrow p=\large\frac{q}{2}$
Hence $B$ is the correct option.
answered Aug 27, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App