Consider a sphere of fixed radius
We want to maximize the volume of the cone,
From the diagram we find the following expression for in terms of
Thus, our expression for
Taking the derivative of this we have
Setting this equal to 0 we have
(We used that
Then, plugging this value of back into our expression for we have
Ok. In the image above, there are two sketches. The one on the left is in 3d, but is kinda hard to refer to. The one on the right is in 2d and easier to see what's going on, so I'm going to use that (just imagine it's a cross-section of the 3d pic).
The secret of the whole problem is to relate $h$ and $b$. This is done using the equation for the right half of the circle: $$x = \sqrt{r^2-y^2}$$
$b$ is the $x$ value when $y$ is offset from the top of the sphere by $h$. Thus, the equation in terms of $b$, $h$, and $r$ is: $$b = \sqrt{r^2 - (r-h)^2}$$
Now to relate to the volume of a cone: $$V_{cone} = \frac\pi 3 \cdot b^2 h$$ $$V_{cone} = \frac\pi 3 \cdot (r^2 - (r - h)^2) h$$ Simplifying: $$V_{cone} = \frac \pi 3 \cdot (2h^2 r-h^3)$$ Differentiate: $$\frac{dV_{cone}}{dh} = \frac \pi 3 \cdot (4r - 3h)h$$ Solve: $$h = \left\{0, \frac{4r}{3}\right\}$$ We want the maximum, which obviously will occur at the second value of $h$. Finding the maximal volume is a simple algebra problem now.