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 $[0,8]$