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 $\displaystyle \frac{4}{\sqrt{6}} \text{ by } \frac{16}{3}$