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I'm trying to understand as to why, with a given volume, the diameter and height of a cylinder need to be the same for the minimum surface area. This thread, shows how to derive the minimum surface area of a cylinder with a given volume, but doesn't explain why this is.

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2 Answers 2

Volume $V=\pi r^2 h$.

Surface area $$S =2 \pi r h +2 \pi r^2$$

Now, the AM-GM inequality says

$$\frac{ \pi r h + \pi rh +2 \pi r^2}{3} \geq \sqrt[3]{\pi r h \cdot \pi rh \cdot2 \pi r^2}= \sqrt[3]{2 \pi V^2 }$$

with equality if and only if $ \pi r h = \pi rh =2 \pi r^2$.

Basically, it boils down to this simple principle (which is more or less AM-GM): if some expressions have a constant product, their sums is minimal if the functions are equal (if they can be equal).

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See here artofproblemsolving.com/Wiki/index.php/Proofs_of_AM-GM for proof of the AM-GM inequality.The proof by Cauchy is the one I really like. –  Amitabh Udayiman May 6 '12 at 18:20
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The thread you linked to shows that making the cylinder very tall or very short increases the surface area, so there must be an optimum somewhere in the middle. It just happens in this problem that the height and diameter are equal. If you did the problem for a can with only one end, as in the linked problem, the answer would change to have the height and radius equal. The direction of the shift is easy to see. If you have only one end, increasing the radius is cheaper in terms of surface area than if you have both ends, so you would increase the radius relative to the height.

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Hmmm...I started editing your typo (that the radius and diameter are equal), but then I got confused by your reference to the two different problems. In brief, the current problem (for a can with both ends) has the height equal to the diameter; and the linked problem (for an open-topped can) has the height equal to the radius. –  TonyK May 6 '12 at 19:52
    
@TonyK: I got confused between the two cases. I believe it is fixed now. –  Ross Millikan May 6 '12 at 21:21
    
That's better! In fact it's obvious, if you think about it, that the optimal shape for the two-ended case has to be two copies of the open-ended case stuck together. –  TonyK May 6 '12 at 23:48
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