Question

The least value of the function $f\left(x\right)=ax+\frac{b}{x}\left(a>0,b>0,x>0\right)$ is

• A. $\sqrt{ab}$
• B. $\frac{2\sqrt{a}}{b}$
• C. $\frac{2\sqrt{b}}{a}$
• D. $2\sqrt{ab}$

Answer 1

The correct option is D

$2\sqrt{ab}$

Explanation for the correct option:

Step 1: Calculate the first derivative and apply extremum condition

The given function is $f\left(x\right)=ax+\frac{b}{x}$, where $a>0,b>0,x>0$.

Differentiating the function $f\left(x\right)$ with respect to $x$,

$f\text{'}\left(x\right)=a-\frac{b}{x^2}$ … $\left[\because \frac{d}{dx}x=1,\frac{d}{dx}{x}^n=n{x}^{n-1}\right]$

Now, as we know for a function $f\left(x\right)$ to be maximum or minimum, it is necessary that the derivative of the function is zero.

i.e., $f\text{'}\left(x\right)=0$

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

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

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

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

$\Rightarrow$ $x=\sqrt{\frac{b}{a}}$ …$\left[\because x>0\right]$

Step 2: Calculate the second derivative and apply extremum condition

Now, on differentiating $f\text{'}\left(x\right)$ with respect to $x$, we get,

$f\text{'}\text{'}\left(x\right)=\frac{2b}{x^3}$ …$[\because \frac{d}{dx}C=0,\frac{d}{dx}{x}^n=n{x}^{n-1}]$

Now, we know that for the function $f\left(x\right)$ to be minimum, it is necessary that the second derivative of the function is greater than zero for the calculated value of $x$.

i.e., $f\text{'}\text{'}\left(x\right)>0$ for $x=\sqrt{\frac{b}{a}}$

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

$\Rightarrow$ $f\text{'}\text{'}\left(x\right)=\frac{2b}{{\left(\sqrt{{\displaystyle \raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}}\right)}^3}$ …$\left[\because x=\sqrt{\frac{b}{a}}\right]$

$\Rightarrow$ $f\text{'}\text{'}\left(x\right)=\frac{2b}{{\displaystyle \raisebox{1ex}{$b\sqrt{b}$}\!\left/ \!\raisebox{-1ex}{$a\sqrt{a}$}\right.}}$

$\Rightarrow$ $f\text{'}\text{'}\left(x\right)=\frac{2a\sqrt{a}}{{\displaystyle \sqrt{b}}}>0$ …$[\because a>0,b>0]$

Thus, the given function $f\left(x\right)$ is minimum at $x=\sqrt{\frac{b}{a}}$.

Step 3: Calculate the minimum value of $f\left(x\right)$

As we have concluded that the given function $f\left(x\right)$ is minimum at $x=\sqrt{\frac{b}{a}}$.

Then, the minimum value of $f\left(x\right)$ is,

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

$\Rightarrow$ $f\left(x\right)=a\times \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{{\displaystyle \raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}}}$ ...$\left[\because x=\sqrt{\frac{b}{a}}\right]$

$\Rightarrow$ $f\left(x\right)=a\times \frac{\sqrt{b}}{\sqrt{a}}+\frac{b\sqrt{a}}{\sqrt{{\displaystyle b}}}$

$\Rightarrow$ $f\left(x\right)=\sqrt{a}\times \sqrt{b}+\sqrt{b}\sqrt{a}$

$\Rightarrow$ $f\left(x\right)=\sqrt{ab}+\sqrt{ab}$

$\Rightarrow$ $f\left(x\right)=2\sqrt{ab}$

Hence, option D is the correct option.