# Maximizing volume of sphere on top of open cone.

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?

• Possible duplicate of Sphere on top of a cone. Maximum volume? Commented Jun 8, 2017 at 7:41
• Yes but you can make the sphere fit 100% of it's volume inside the cone. However that is not nessecarily the maximum volume that the sphere can occupy from the cone. I need to maximize $V_{cap}/V_{cone}.$ Commented Jun 8, 2017 at 7:58
• It's quite clear: math.stackexchange.com/questions/2308368/… Also, in my comment to the answer below. Commented Jun 8, 2017 at 8:02

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.

• How did you get that $$AH=\frac{\cos^2{\alpha}}{\sin{\alpha}}?$$ And what do you mean by "$f'(x)$ vanishes..." ? Commented Jun 8, 2017 at 15:07
• $AE=1/\tan\alpha=\cos\alpha/\sin\alpha$ and $AH=AE\cdot\cos\alpha$. By "$f'(x)$ vanishes" I mean $f'(x)=0$, where $f'(x)$ is the derivative of $f(x)$. Commented Jun 8, 2017 at 16:22
• Solving $f'(x)=0$ for $x$ I get that $x=\sin\alpha -1$ and not what you got. I did this with Maple. Commented Jun 8, 2017 at 18:32
• I also don't get how you get the lenght of $DF$ to $1-\sin{\alpha}+x.$ I get that $DF = AD-AF$, but $AF$ is still unknown. Commented Jun 8, 2017 at 18:52
• I'm afraid Maple lost a solution: $f'(x)=0$ reduces to $x^2=1+\sin^2\alpha-2\sin\alpha$, that is $x=\pm(1-\sin\alpha)$. But $x$ must be positive, so the negative solution has to be discarded. Commented Jun 8, 2017 at 19:54

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.

• So, $k$ becomes $$k=r\cdot \frac{1}{\tan\alpha}-\frac{1}{\sin\alpha}+1$$? Commented Jun 8, 2017 at 7:08
• I think so, yes. Commented Jun 8, 2017 at 7:15
• This seems wrong. I get that $V_{cap}/V_{cone}$ is $$f(\alpha) = \frac{\pi k^2(3r-k)/3}{\pi r^2 \cdot \frac{r}{\tan\alpha}/3}$$ which gets eve more intimidating if I plug in $k$ and has no maximum. Commented Jun 8, 2017 at 7:15