Question

What is the minimum value of the function `y=sqrt(x²+2ax+10a²)` where a > 0 ?

Answer 1

In order to find the minimum value of the function we have to minimize `g(x)=x^2+2ax+10a^2` which occurs at

`g'(x)=0=>2x+2a=0=>x=-a`

and `g(-a)=9a^2`

Hence `f(-a)=sqrt(g(-a))=sqrt(9a^2)=3absa=3a` since `a>0`