TITLE: Sphere on top of a cone. Maximum volume?
QUESTION [5 upvotes]: I received an interesting problem which I can not figure how to solve. It follows: An ice cream has the shape of a sphere and a cone like the image below. What is the maximum volume that the sphere can occupy out of the cone? Note that $M$ is the center of the sphere, not the center of the circle of the base of the cone. I hope you understand my phrasing. 
A rephrasing of the problem statement would be, I believe:  Say you have an arbitrary cone, and you are going to place a sphere on it with radius $r$, what is the maximum volume of the cone the sphere can occupy, provided some part of the sphere-cap is allowed to be above the circular cone base?
This is my drawing of the problem:

Attempt so far: I know that $\bigtriangleup AEM\sim\bigtriangleup ACG$. Denote $AM=x$, $GH=h$, then $GM=h-r.$ By the Pythagoran theorem on $\bigtriangleup ACG$ I get that 
$$AC=\sqrt{(x+h-r)^2+R^2}.$$
Since $R/AC=k$ is a constant, by similar triangles I obtain 
$$\frac{r}{x}=\frac{R}{AC}=k \Leftrightarrow r=kx$$ 
The volume of the cone is
$$V_{cone}=\frac{R^2\pi(x+h-r)}{3}=\frac{R^2\pi(x+h-kx)}{3}.$$ 
Using the formula for a spherical cap, I get that the volume of the sphere that is below the cone base is 
$$V_{s.cap}=\frac{\pi h^2(3r-h)}{3}.$$ 
Thus the ratio is 
$$f(x)=\frac{V_{s.cap}}{V_{cone}}=\frac{\frac{\pi h^2(3r-h)}{3}}{\frac{R^2\pi(x+h-kx)}{3}} = \frac{h^2(3r-h)}{R^2(h+(1-k)x)}.$$
But this does not work.

REPLY [3 votes]: Let's suppose first of all the cone is given, with base radius $R$ and height $H$, and set $x=r/R$, $t=H/R$.
By similar triangles we have 
$\displaystyle MA={r\over R}AC={r\over R}\sqrt{R^2+H^2}=r\sqrt{1+t^2}$
and the height $h$ of the spherical cap inside the cone can then be written as
$$
h=r+H-MA=H-r\left(\sqrt{1+t^2}-1\right).
$$
Notice that this makes sense as long as $0\le h\le 2r$, which entails:
$$
{\sqrt{1+t^2}-1\over t}\le x \le {\sqrt{1+t^2}+1\over t}.
$$
The ratio between the volume of the spherical cap inside the cone and the volume of the cone can then be written as:
$$
f(x)=\frac{V_{s.cap}}{V_{cone}}= \frac{h^2(3r-h)}{R^2H}
={1\over t}\left(t+x-x\sqrt{1+t^2}\right)^2\left(2x-t+x\sqrt{1+t^2}\right),
$$
where $t$ is fixed and $x$ can vary between the limits given above.
By differentiating this it turns out that $f(x)$ has two stationary points: a local minimum at $x={\sqrt{1+t^2}+1\over t}$ (which is the upper bound for $x$) and a local maximum at
$$
x_\max={t\sqrt{1+t^2}\over \sqrt{1+t^2}-1+t^2},
$$
which is inside the permitted range for $x$.
Plugging this into $f(x)$ we can find the maximum volume ratio $f_\max$ as a function of $t$:
$$
f_\max(t)=\frac{4 \left(\sqrt{t^2+1}(1+t^2)-3t^2+1\right)}
{\left(t^2-3\right)^2}.
$$
By differentiating one finds
$$
f'_\max(t)=-{4t\over\big(2+\sqrt{1+t^2}\big)^3},
$$
which is negative for $t>0$.
It follows that $f_\max(t)$ is a monotonically decreasing function for $t>0$, thus 
$$
f_\max(t)<f_\max(0)={8\over9},
\quad\hbox{for $t>0$}.
$$ 
But of course $t=0$ is a degenerate case, because both volumes vanish as $t\to0$. Hence this maximum value should be regarded as a limiting case: the ratio $V_{s.cap}/V_{cone}$ can never reach $8/9$ but can be as close to it as one wants.