# What are the absolute extrema of f(x)=x^(1/3)*(20-x)in[0,20]?

Feb 14, 2018

The absolute minimum is $0$, which occurs at $x = 0$ and $x = 20$.

The absolute maximum is $15 \sqrt[3]{5}$, which occurs at $x = 5$.

#### Explanation:

The possible points that could be absolute extrema are:

1. Turning points; i.e. points where $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

2. The endpoints of the interval

We already have our endpoints ($0$ and $20$), so let's find our turning points:

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

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{3}} \left(20 - x\right)\right) = 0$

$\frac{1}{3} {x}^{- \frac{2}{3}} \left(20 - x\right) - {x}^{\frac{1}{3}} = 0$

$\frac{20 - x}{3 {x}^{\frac{2}{3}}} = {x}^{\frac{1}{3}}$

$\frac{20 - x}{3 x} = 1$

$20 - x = 3 x$

$20 = 4 x$

$5 = x$

So there is a turning point where $x = 5$. This means that the 3 possible points that could be extrema are:

$x = 0 \text{ "" "x=5" "" } x = 20$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's plug these values into $f \left(x\right)$:

$f \left(0\right) = {\left(0\right)}^{\frac{1}{3}} \left(20 - 0\right) = 0 \cdot 20 = \textcolor{red}{0}$

f(5) = (5)^(1/3) (20 - 5) = root(3)(5) * 15 = color(red)(15root(3)5

$f \left(20\right) = {\left(20\right)}^{\frac{1}{3}} \left(20 - 20\right) = \sqrt[3]{20} \cdot 0 = \textcolor{red}{0}$

Therefore, on the interval $x \in \left[0 , 20\right]$:

The absolute minimum is $\textcolor{red}{0}$, which occurs at $x = 0$ and $x = 20$.

The absolute maximum is $\textcolor{red}{15 \sqrt[3]{5}}$, which occurs at $x = 5$.