Maximizing volume of sphere on top of open cone.

Question

How do I maximize the volume of the part of the sphere that is below the plane $BC$? I need to maximize the ratio $V_{sphere cap}/V_{cone}.$

I will work with this problem in 2D first. Please observe this figure:

Observe that $AD=h.$ In order to use the formula for spherical cap, which is $$V_{cap}=\frac{\pi k^2(3r-k)}{3}, \ \ \ \ \ \text{where} \ k=DM+1.$$So $BA = \sqrt{r^2+h^2}.$ Thus, since $\bigtriangleup ABD \sim \bigtriangleup AEM,$ I get

$$\frac{r}{\sqrt{r^2+h^2}}=\frac{1}{AM} \Longrightarrow AM=\frac{\sqrt{r^2+h^2}}{r}.$$

However, I also know that

$$\sin{\alpha} = \frac{r}{\sqrt{r^2+h^2}} \Longrightarrow \frac{1}{\sin\alpha}=\frac{\sqrt{r^2+h^2}}{r}=AM$$

This means that $$DM=h-AM=h-\frac{\sqrt{r^2+h^2}}{r} = h - \frac{1}{\sin\alpha}$$

and $$k=DM+1=h+1-\frac{1}{\sin{\alpha}}.$$

I also know that $h=r/\tan{\alpha}$ but I still get a function with 2 variables. any idea on how to proceed?

Answer 1

Let angle $\alpha$ be fixed: that part of the cone between vertex and tangency circle is then also fixed. If $H$ is the projection of $E$ onto $AD$ in your diagram, then $\displaystyle AH={\cos^2\alpha\over\sin\alpha}$.

Set now $DH=x$. Total height of the cone is of course $\displaystyle AD={\cos^2\alpha\over\sin\alpha}+x$, while the height of the spherical cap is $DF=1-\sin\alpha+x$. Notice that this makes sense as long as $DF\le2$, that is if $x\le1+\sin\alpha$. From these we can easily find an expression for the volume ratio: $$ f(x)=\frac{V_{s.cap}}{V_{cone}}= \frac{DF^2(3-DF)}{AD^3\tan^2\alpha} =\frac{(1-\sin\alpha+x)^2(2+\sin\alpha-x)}{\big({\cos^2\alpha\over\sin\alpha}+x\big)^3\tan^2\alpha}. $$ $f'(x)$ vanishes only for $x=1-\sin\alpha$, which is the maximum point. Plugging that into the expression for $f(x)$ we get its maximum value: $$ f_\max=\frac{4 (1-\sin\alpha)^2 (1+2 \sin\alpha) \cot ^2\alpha} {(1-\sin\alpha+\cos^2\alpha/\sin\alpha)^3}. $$ This is a function of $\alpha$, attaining its maximum (which is ${8/9}$) for $\alpha=\pi/2$. This corresponds with the result already found here.

Notice however that for $\alpha=\pi/2$ the cone degenerates to a point, so this result must be understood as a limiting value, which can be approached as close as one wants, but never reached.

Answer 2

I think $r$ should be fixed because - if I see this right - the volume you want to maximize can get arbitrarily large as $r$ gets larger. Now since we have seen that $r$ is fixed, your function has only one variable, namely $\alpha $. Then you can proceed with standard calculus and take the derivative with respect to $\alpha $. Hope this helps.