Example: A rectangle has its vertices on
the $x$-axis, on the $y$-axis, at the origin, and somewhere on the
graph $y=4-x^2$ in the first quadrant. Find the maximum possible area
of such a rectangle. Justify your answer.

Solution:
As shown in the video, our rectangle has width $x$
and height $y$, and so has area $xy$. But $y=4-x^2$, so our area is
$$A(x)= x(4-x^2) = 4x-x^3.$$ This turns
our story problem into just finding the maximum value
of $A(x)$ on $[0,2]$. Since $A'(x)=4-3x^2$, we have a critical number
when $4-3x^2=0$, or $x = \frac{2}{\sqrt{3}}$. Then
$y= 4-x^2 = 4 - \frac{4}{3} = \frac{8}{3}$, and the area is $$A =
xy = \frac{16}{3\sqrt{3}} = \frac{16 \sqrt{3}}{9}.$$

(The video cuts out a few seconds too soon, but the end of the
calculation is shown to the left.)