# Find the least value

Question:

Find the least value of $\mathrm{f}(\mathrm{x})=a x+\frac{b}{x}$, where $\mathrm{a}>0, \mathrm{~b}>0$ and $\mathrm{x}>0$.

Solution:

We have,

$f(x)=a x+\frac{b}{x}$

$\Rightarrow f^{\prime}(x)=a-\frac{b}{x^{2}}$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow a-\frac{b}{x^{2}}=0$

$\Rightarrow x^{2}=\frac{b}{a}$

$\Rightarrow x=\sqrt{\frac{b}{a}},-\sqrt{\frac{b}{a}}$

But, $x>0$

$\Rightarrow x=\sqrt{\frac{b}{a}}$

Now,

$f^{\prime \prime}(x)=\frac{2 b}{x^{3}}$

At $x=\sqrt{\frac{b}{a}}$

$f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}}=\frac{2 a^{\frac{3}{2}}}{b^{\frac{1}{2}}}>0 \quad[\because a>0$ and $b>0]$

So, $x=\sqrt{\frac{b}{a}}$ is a point of local minimum.

Hence, the least value is

$f\left(\sqrt{\frac{b}{a}}\right)=a \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}$