Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 24

Determine the dimensions of the rectangle of largest area that has its base on the $x$-axis and its other two vertices above the $x$-axis and lying on the parabola $y = 8 - x^2$




The Area of the rectangle is $A = 2x(y) = 2xy$
$A' = 16 - 6x^2$
By Taking the derivative...
$A' = 16-6x^2$
when $A' = 0$,

$
\begin{equation}
\begin{aligned}
0 &= 16 - 6x^2\\
\\
x^2 &= \frac{16}{6}\\
\\
x &= \sqrt{\frac{16}{6}} = \frac{4}{\sqrt{6 }}
\end{aligned}
\end{equation}
$


when $A' = 0$,

$
\begin{equation}
\begin{aligned}
0 &= 16 - 6x^2\\
\\
x^2 &= \frac{16}{6}\\
\\
x &= \sqrt{\frac{16}{6}} = \frac{4}{\sqrt{6}}
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{if } x &= \frac{4}{\sqrt{16}}, \text{ then}\\
\\
y &= 8 - x^2 = 8 - \left( \frac{4}{\sqrt{6}} \right)^2\\
\\
y &= \frac{16}{3}
\end{aligned}
\end{equation}
$


Therefore, the dimensions of the rectangle is $\displaystyle \frac{4}{\sqrt{6}} \text{ by } \frac{16}{3}$