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Lets say we have a sphere, cube, and cylinder of same dimensions, that is The diameter of the sphere is same as one side of a cube and same as circumference of the cylinder. Of these three containers, how can one prove that the sphere has largest capacity?

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closed as unclear what you're asking by amWhy, le gâteau au fromage, hardmath, Davide Giraudo, Claude Leibovici Sep 7 at 16:07

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What is the height of the cylinder? –  workmad3 Jul 28 '10 at 13:15
if the diameter of the sphere is same as one side of the cube, the sphere is inscribed in the cube. Are you sure you don't mean the radius of the sphere? (and can you vouch it's not a homework question?) –  mau Jul 28 '10 at 13:18

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up vote 4 down vote accepted

It is easy to show which has the largest volume (once you've provided the height of the cylinder as per my comment) as long as you accept proofs built on top of the existing verified equations of the volumes of each solid. Then you have

$sphere = \frac{4\pi}{3}(\frac{d}{2})^3 = \frac{\pi d^3}{6}$

$cube = d^3$

$cylinder = (\frac{d}{2\pi})^2 \pi h = \frac{d^2}{4\pi^2}h\pi = \frac{hd^2}{4\pi}$

The final equation would be: $\frac{d^3}{4\pi}$ if you make the height of the cylinder equal to it's circumference and therefore equal to the diameter of the sphere.

This actually shows a different result to the one you suggested too, namely that the cube has the greatest volume in this setup. This can easily be seen geometrically by envisioning the shapes. The sphere with diameter $d$ will fit inside a cube with side of length $d$. The cylinder with circumference $d$ is the smallest (assuming the height is equal to the circumference) because it is a cylinder with diameter $\frac{d}{2\pi}$ which is a much smaller circle than that described by a cross-section of the sphere taken through a great circle drawn on the surface.

I think you may have gotten your assumption from a related but different statement - namely that a sphere has the largest volume given that the surface areas are equal in the shapes being considered. You can show this in a similar way by simply comparing the equations for volumes again, but calculating them from the surface area equations instead of directly from sides/diameters.

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Consider making your answers a little less complete on HW-style questions. –  BBischof Jul 28 '10 at 17:17
@BBishof: True, I would have just given links to Wikipedia articles –  Casebash Jul 28 '10 at 21:35

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