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Largest cone inside a sphere

Anonymous Thu 13 Jan 20:56:53 2022 No.14098248 View Reply Original Report

Quoted By: >>14098447 >>14099027 >>14099437

I have a sphere $S$ with a radius $R$ . What is the maximum lateral area of a cone inscribed in $S$ ?

I solved this problem with very basic diff geometry - we have a trivial parametrization thus a trivial way to compute the $\mathrm{d} \sigma$ but it seems like I am using something that is too much sophisticated.

Anonymous

Anonymous Thu 13 Jan 2022 21:28:39 No.14098409 Report

Quoted By:

>largest cone you can fit inside spheres? i
idk ask ur mom.

Anonymous

Anonymous Thu 13 Jan 2022 21:37:36 No.14098447 Report

Quoted By:

>>14098248
Did you get a=R/3? I think the way I did it was totally elementary, although I assumed the only possible cones were inscribed and right circular as in your picture.

Anonymous

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Anonymous Thu 13 Jan 2022 23:35:09 No.14099027 Report

Quoted By: >>14099072 >>14099099

>>14098248
After consultation with our business unit we determined that you have one (1) free question remaining after which your credit card will be automatically charged. Thank you for shopping with 4chan.

The volume of a cone is $r^2 \cdot \pi\frac{h}{3}$ . Imagine that the cone lies in the blue circle f(x), with its tip at the origin. Then, the height of the cone is x and its radius f(x). Therefor, its volume becomes v(x), in thick orange. In thin blue, right on top of the same line, is a simplification of v(x) after you substitute f(x) into it. It shows us we substituted correctly and the same are identical. Then, find the maximum of v(x) by finding a zero of its derivative v'(x). That is at $\frac{4}{3}r$ .

Anonymous

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Anonymous Thu 13 Jan 2022 23:46:20 No.14099072 Report

Quoted By: >>14099079 >>14099085

>>14099027
I changed a few cosmetics. The solution is still the same, but it's a whole circle with the cone's outline traced in red.

Anonymous

Anonymous Thu 13 Jan 2022 23:48:20 No.14099079 Report

Quoted By:

>>14099072
I didn't realize I could share my graph without an account. Apparently I can, so here is a link where you can adjust the radius r, using the slider.
https://www.desmos.com/calculator/0exg9qn01a

Anonymous

Anonymous Thu 13 Jan 2022 23:49:39 No.14099085 Report

Quoted By: >>14099091

>>14099072
Beautiful work anon. Link to the desmos graph please?

Anonymous

Anonymous Thu 13 Jan 2022 23:51:03 No.14099091 Report

Quoted By:

>>14099085
nvm, just saw the post above mine

Anonymous

Anonymous Thu 13 Jan 2022 23:52:15 No.14099099 Report

Quoted By: >>14099140

>>14099027
I don't like that you seemed to just assume maximizing the volume should maximize the lateral surface area. You should either show this or just write out the formula for the lateral surface area and maximize that instead.

Anonymous

Anonymous Fri 14 Jan 2022 00:02:04 No.14099140 Report

Quoted By: >>14099155

>>14099099
I don't think I assumed that. v(x) represents ALL possible volumes, depending on height x. The radius follows for any height x, because we want it to be maximal: at any given height there is no point in making the cone's radius any less than f(x), because it is apparent that we would be wasting volume by doing so. It does not what point on the sphere we use for the location of the cone's tip, any angling away of the cone's axis from the chord passing through the cone's tip and the center of the sphere would reduce the volume of the cone, for the given height.

You are right that I had left this implied and I hope that it's sufficient as a very informal proof.

Anonymous

Anonymous Fri 14 Jan 2022 00:06:33 No.14099155 Report

Quoted By: >>14099157

>>14099140
You did assume that, and you didn't even need to reference the volume at all, just use that LSA=pi*r*L, write L in terms of r and x, differentiate, and then set equal to 0 and solve using the quadratic formula.

Anonymous

Anonymous Fri 14 Jan 2022 00:07:24 No.14099157 Report

Quoted By: >>14099349

>>14099155
Actually, I had misread the question. He did ask about the lateral surface area, not the volume. You are right that it's not obvious. I will check if the solution is still correct. The approach might be very similar.

Anonymous

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Anonymous Fri 14 Jan 2022 00:45:26 No.14099349 Report

Quoted By:

>>14099157
The lateral surface area of a cone is $\pi r\sqrt{h^{2}+r^{2}}$ .

The same approach can be used as for the volume. It works out to be the same, so the height at which we maximize the volume also maximizes the lateral surface. I admit that I had not planned it like this and really just misread the question, but in hindsight it almost seems as if this would have been the easier solution: the equations are quite a bit simpler in the volume case. But, the other anon is right that we should first have made a proper argument that the two coincide. With that argument out of the way, even if someone had not made my mistake, it would have maybe been a good conscious choice to do so, in order to work with simpler equations.

$v_{max}$ is now as shown in the screenshot. The link to the updated graph:
https://www.desmos.com/calculator/3gsynlr7wa

I hope most mistakes are gone now.

Anonymous

Anonymous Fri 14 Jan 2022 01:07:08 No.14099437 Report

Quoted By: >>14099479

>>14098248
I need to pay more attention, OP. So, you had already solved it. But, I suppose this is still valuable for you to double check your approach and for other anons to see how one could do it.

But you wanted to see if there is an easier way? The closed form, simplified form of the lateral surface, as shown in the graphs, is probably hard to get to by any other way, but I wonder, which might be the same reason for you posting your question, if there is some simpler reasoning to get to $\frac{4}{3}r$ . From there, it's really only substitution, so if someone could make a good argument, maybe not using derivatives to find the maximum, why that height should maximize the lateral area / volume, then one could work with easier tools and possibly skip the use of the derivative. I'll think about that more.

Anonymous

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Anonymous Fri 14 Jan 2022 01:13:45 No.14099479 Report

Quoted By: >>14099515

>>14099437
Here is a solution from Stack Exchange, but it's for the volume problem again, the question I had mistaken it as at first. But if someone can come up with an argument not using derivatives for the volume, because of the argument above, one can probably also come up with an argument not using derivatives for the lateral surface problem. He has a slightly different approach, but still ends up arguing about the maximum with derivatives.

I'll keep looking for some other, simpler, better argument for why it is at 4/3r. My intuition tells me that this fraction is just a little too nice that one can't find a different argument for it.

https://math.stackexchange.com/questions/1176722/max-volume-of-a-cone-in-a-sphere

Anonymous

Anonymous Fri 14 Jan 2022 01:20:11 No.14099515 Report

Quoted By: >>14099743

>>14099479
Here are some other solutions. Some of them are more complicated, but all of them use the derivative somehow.
https://math.stackexchange.com/questions/225718/find-the-volume-of-the-largest-right-circular-cone-that-can-be-inscribed-in-a-sp/225731
https://math.stackexchange.com/questions/3502006/maximize-volume-surface-of-a-cone-that-fits-in-a-sphere
https://math.stackexchange.com/questions/2698088/maximize-the-volume-of-a-cone

Now that I've started on it, I really want to find an argument without derivatives, too. Why 4/3r? There must be some argument to get to 4/3 by some geometric argument.

Anonymous

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Anonymous Fri 14 Jan 2022 02:00:47 No.14099743 Report

Quoted By: >>14099845

>>14099515
I'm not sure if this will be fruitful, but I wonder if Cavalieri's principle can somehow be used to reason about the 4/3r for the height of the cone.
https://en.wikipedia.org/wiki/Cavalieri%27s_principle#Spheres

There seem to be very few sources, or none, on the Internet that make such an argument. I have looked through dozens of videos, blogs, and Q&A sites now. They all argue with the derivative, so that might make it more likely that if someone could make a geometric argument with something like Cavalieri's principle or something really out of the ordinary, that might actually be something interesting.

Anonymous

Anonymous Fri 14 Jan 2022 02:16:28 No.14099845 Report

Quoted By:

>>14099743
Actually, I think I got something! I need to clean it up a bit. The reasoning works from the cylinder inscribed into a sphere, to the relationship between the volume of the cone inscribed into the cylinder to the cylinder, to the volume of the cone, to the relationship between the volume of the cone to the lateral surface of the cone. If you put it all together, I think you can make an argument that it must be 3/4 of the radius without any further math at all. That would be kind of cool, IMO. I'll get back to you.

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