Answer
$V = \frac{32\pi R^{3}}{81}$
Work Step by Step
Using the sketch, we can see that the radius of the cone $r$ , $x$, and $R$ form a right triangle. Therefore $x^{2}$ + $r^{2}$ = $R^{2}$
We can solve the equation for $r^2$
$r^{2}$ = $R^{2}- x^{2}$.
The height of the cone is $h = x + R$
We know that the volume of a cone is $V = \frac{1}{3}\pi r^{2}h$. So if we plug the 2 values in we get volume in terms of an unknown, $x$ and a constant $R$.
$V = \frac{1}{3}\pi r^{2}h$
$V = \frac{1}{3}\pi (R^{2}$ - $x^{2})(x + R)$
Use the product rule to find $\frac{dV}{dx}$
$\frac{dV}{dx} = \frac{1}{3}\pi ((-2x)(R+x) + (R^{2}$ - $x^{2}))$
$\frac{dV}{dx} = \frac{1}{3}\pi (-2xR-2x^2+R^2-x^2)$
Set $\frac{dV}{dx} = 0$ , to find maximum volume
$0 = \frac{1}{3}\pi (-2xR-2x^2+R^2-x^2)$
$0 = -3x^2 - 2xR + R^2$
$0 = 3x^2 + 2xR - R^2$
$0 = (3x-R)(x+R)$
$x = \frac{R}{3}$ and $x =-R$
Using sign analysis we find that $x =-R$ is a local min of $\frac{dV}{dx}$ and $x = \frac{R}{3}$ is a local max.
Alternatively, we can reason: if $x = -R$, and $h = x + R$
$h = -R + R$
$h = 0$
$V = \frac{1}{3}\pi r^{2}(0)$
$V = 0$.
We can plug $x$ back into the original equations to find $h$ and $r$.
$h = R+x$
$h = R + (\frac{R}{3})$
$h= \frac{4}{3}R$
$r^2 = R^2 - x^2$
$r^2 = R^2 - (\frac{R}{3})^2$
$r^2 = \frac{8}{9}R^2$
Plug $r$ and $h$ back in to find $V$
$V = \frac{1}{3}\pi r^2h$
$V = \frac{1}{3}\pi (\frac{8}{9}R^2)(\frac{4}{3}R)$
$V = \frac{32\pi R^{3}}{81}$
