Question

The minimum value of f(x) = x² + $\frac{250}{x}$ is _______________.

Answer 1

The given function is $f\left(x\right)=x^2+\frac{250}{x}$, x ≠ 0.

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

Differentiating both sides with respect to x, we get

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

For maxima or minima,

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

$\Rightarrow 2x-\frac{250}{x^2}=0$

$\Rightarrow x^3=\frac{250}{2}=125$

$\Rightarrow x=5$

Now,

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

At x = 5, we have

$f\text{'}\text{'}\left(5\right)=$$2+\frac{500}{{\left(5\right)}^3}=$$2+\frac{500}{125}=$2 + 4 = 6 > 0

So, x = 5 is the point of local minimum.

∴ Minimum value of f(x) = $f\left(5\right)={\left(5\right)}^2+\frac{250}{5}$ = 25 + 50 = 75 $\left[f\left(x\right)=x^2+\frac{250}{x}\right]$

Thus, the minimum value of the given function is 75.


The minimum value of $f\left(x\right)=x^2+\frac{250}{x}$ is ____75____.