Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 54

Determine the absolute maximum and absolute minimum values of $\displaystyle f(x) = \sqrt[3]{x} (8-x)$ on the interval $[0,8]$.

Simplifying the equation,

$
\begin{equation}
\begin{aligned}
f(x) &= x^{\frac{1}{3}} (8-x)\\
\\
f(x) &= 8x^{\frac{1}{3}} - x^{\frac{4}{3}}
\end{aligned}
\end{equation}
$


Taking the derivative of $f(x)$, we have...

$
\begin{equation}
\begin{aligned}
f'(x) &= 8 \left( \frac{1}{3} \right) x^{\frac{-2}{3}} - \frac{4}{3} x^{\frac{1}{3}}\\
\\
f'(x) &= \frac{4}{3} \left( 2x^{\frac{-2}{3}} - x^{\frac{1}{3}}\right)
\end{aligned}
\end{equation}
$



Solving for critical numbers, when $f'(x) = 0$
$0 = \frac{4}{3}\left( 2x^{\frac{-2}{3}} - x^{\frac{1}{3}} \right)$


$
\begin{equation}
\begin{aligned}
0 &= 2x^{\frac{-2}{3}} - x^{\frac{1}{3}}\\
\\
2x^{\frac{-2}{3}} &= x^{\frac{1}{3}}\\
\\
2 &= x^{\frac{1}{3}} \left( x^{\frac{2}{3}} \right) \\
\\
x^{\frac{3}{3}} &= 2\\
\\
x &= 2
\end{aligned}
\end{equation}
$



We have either absolute maximum and minimum values at $x = 2$
So,

$
\begin{equation}
\begin{aligned}
\text{when } x &= 2\\
\\
f(2) &= \sqrt[3]{2} (8-2)\\
\\
f(2) &= 7.5595
\end{aligned}
\end{equation}
$

Evaluating $f(x)$ at end points $x = 0$ and $x = 8$, we have...

$
\begin{equation}
\begin{aligned}
\text{when } x &= 0,\\
\\
f(0) &= \sqrt[3]{0} (8-0)\\
\\
f(0) &= 0
\\
\\
\\
\text{when } x &= 8\\
\\
f(8) &= \sqrt[3]{8} (8-8)\\
\\
f(8) &= 0
\end{aligned}
\end{equation}
$

Therefore, we have absolute maximum value at $\displaystyle f(2) = 7.5595$ and the absolute minimum value at $f(0) = f(5) = 0 $ on the interval $[0,8]$