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### Answers (1)

Answer:

$2\sqrt{ab}$

Hint:

For maxima or minima we must have ${f}'\left ( x \right )=0$

Given:

$f\left ( x \right )=ax+\frac{b}{x}$                                   $a> 0,b> 0,x> 0$

Solution:

$f\left ( x \right )=ax+\frac{b}{x}$

${f}'\left ( x \right )=a-\frac{b}{x^{2}}$

For critical points

${f}'\left ( x \right )=0$

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

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

$x=\underline{+}\sqrt{\frac{b}{a}}$

But $x> 0$

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

${{f}'}'\left ( x \right )=-b\left ( \frac{-2}{x^{3}} \right )$

${{f}'}'\left ( x \right )=\frac{2b}{x^{3}}$

${{f}'}'\ \left ( \sqrt{\frac{b}{a}} \right ) =\frac{2b}{\left ( \sqrt{\frac{a}{b}} \right )^{3}}> 0$ as $a> 0,b> 0$

At $x=\sqrt{\frac{b}{a}}$  we will get minimum value

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

$=\sqrt{ab}+\sqrt{ab}$

$=2\sqrt{ab}$

$\therefore$ The least value of $f\left ( x \right )$  is $=2\sqrt{ab}$

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