## Maximum Area of Rectangle Between Parabola and x Axis

What is the maximum area of a rectangle enclosed between the parabola
$y=16-x^2$
and the
$x$
axis?
From the diagram, the base of the rectangle is
$3x$
and the height is
$16-x^2$
os the area is
$A=x(16-x^2)=16x-x^3$
.
$\frac{dA}{dx}=16-3x^2$
.
Set
$\frac{dA}{dx}=0$
and solve to give
$x= \frac{4 \sqrt{3}}{3}$
.
The area is then
$A=16 \frac{4 \sqrt{3}}{3} - (\frac{4 \sqrt{3}}{3})^3= \frac{128 \sqrt{3}}{9}$
.