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Questions very similar to this one have been asked, but I was wondering if it was possible to solve it without calculus, in how some other variants of this question are.

One such example is finding the largest volume of a cube inscribed in a sphere of radius r, where you can use the fact that such a cube would have a diagonal length equal to the diameter, and this would allow you to use the pythagorean theorem, so on and so forth.

I wanted to know whether or not it you can solve the titular problem in the same way; without calculus, only using geometric properties of the shapes.

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  • $\begingroup$ @RicardoCavalcanti Ahhh, I apologize, I meant inscribed in a sphere. Changing now $\endgroup$
    – dfish
    Apr 24, 2021 at 0:40
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    $\begingroup$ Maximising $\frac13\pi h^2(2r-h)$ for $0<h<2r$ is a very simple application of AM-GM. $\endgroup$ Apr 24, 2021 at 0:43

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Draw the cross-section of the sphere and the cone as described here. Marking the centre and the radii, by Pythagoras $(h-r)^2 + b^2 = r^2$ where $b$ is the length of the cone's radius. We do not need to find $b$, just $b^2$, and so this gives $V = \frac{1}{3} \pi b^2 = \frac{1}{3} \pi (r^2 - (h-r)^2) = \frac{1}{3} \pi h^2 (2r- h)$.

Then to maximise $V = \frac{1}{3} \pi h^2 (2r - h)$:

$$\sqrt[3]{\frac{1}{2}h \cdot \frac{1}{2}h \cdot (2r - h)} ≤ \frac{2r}{3} \Rightarrow \frac{1}{4} h^2 (2r - h) ≤ \frac{8r^3}{27}$$ $$\Rightarrow \frac{1}{3} \pi h^2 (2r - h) ≤\frac{32\pi r^3}{81}$$

and we can check equality is satisfied when $\frac{1}{2}h = \frac{1}{2}h = (2r - h)$ or $h = \frac{4}{3}r$. When this is substituted into $V$, $\frac{1}{3} \pi (\frac{4}{3}r)^2 (2r - \frac{4}{3}r) = \frac{32 \pi r^3}{81}$, which proves that we have achieved the best upper bound (this is not guaranteed).

Note that the decomposition of $h \cdot h \cdot (2r - h) \cdot (2r - h)$ does not work as the volume depends on $r^4$ instead of $r^3$.

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  • $\begingroup$ Why in the pythagorean theorem is it (h-r) and not (r-h)? $\endgroup$
    – dfish
    Apr 24, 2021 at 1:53
  • $\begingroup$ It can actually be both of those depending on if $r < h$ or if $r > h$. They both result in the same value as $(h - r)^2 = (-1)^2 (h-r)^2 = (-(h-r))^2 = (r - h)^2$. $\endgroup$
    – Toby Mak
    Apr 24, 2021 at 1:55
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    $\begingroup$ Ok thank you, that's what I thought $\endgroup$
    – dfish
    Apr 24, 2021 at 1:55

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