Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question  7

Answer:  $2 \sqrt{a b}$

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

Solution:

\begin{aligned} &f(x)=a x+\frac{b}{x} \\\\ &f^{\prime}(x)=a-\frac{b}{x^{2}} \end{aligned}

For a local maxima or a local minima we must have

\begin{aligned} &f^{\prime}(x)=0 \\\\ &a-\frac{b}{x^{2}}=0 \\\\ &x^{2}=\frac{b}{a} \end{aligned}

\begin{aligned} &x=\sqrt{\frac{b}{a}},-\sqrt{\frac{b}{a}} \\\\ &\text { But } x>0 \\\\ &x=\sqrt{\frac{b}{a}} \end{aligned}

Now

\begin{aligned} &f^{\prime \prime}(x)=\frac{2 b}{x^{3}} \\\\ &\text { At } x=\sqrt{\frac{b}{a}} \end{aligned}

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

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

Hence the least value is

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