Question

The minimum value of $f\left(x\right)=x^2+\frac{250}{x}$ is

• A. $15$
• B. $25$
• C. $45$
• D. $75$

Answer 1

The correct option is D $75$

Given $f\left(x\right)=x^2+\frac{250}{x}$

$f^{\prime }\left(x\right)=2x-\frac{250}{x^2}$

$=\frac{2(x^3-125)}{x^2}$

$=\frac{2(x^3-5^3)}{x^2}$

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

$\Rightarrow \frac{2(x^3-5^3)}{x^2}=0$

$\Rightarrow x=5$

$f^{\prime \prime }\left(x\right)=2+\frac{500}{x^3}$

$f^{\prime \prime }\left(5\right)=2+\frac{500}{5^3}>0$

Hence $f\left(x\right)$ has minimum value at $x=5$

$f\left(5\right)=25+50$

$\therefore$ the minimum value of $f\left(x\right)=75$