Question

If p,q,r satisfy 4p+3q+r=7 then the least value of 2p^2+q^2+r^2 is

Answer 1

Answer:

The least value of  $2p^2+q^2+r^2$ is  5

Step-by-step explanation:

Given condition is 4p+3q+r=7 we have to find the least value of p, q and r which satisfy the given condition.

Let the value of p,q and r starting from 0

Let p=1, q=1 and r=0 so that these values satisfy the given condition 4p+3q+r=7 and also gives the least value of $2p^2+q^2+r^2$

4(1)+3(1)+0=7 ,

Hence, the least value of  $2p^2+q^2+r^2$ is  $2(1)^2+1^2+0^2$=5