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Given a fixed volume for a solid cylinder, is it possible to find the minimum or maximum curved surface area $ 2 \pi r h $ of this cylinder?

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  • $\begingroup$ No. Why do you ask? $\endgroup$ – David K Dec 4 '14 at 13:48
  • $\begingroup$ Someone asked why is it that one can't find the minimum CURVED surface area but one CAN find the minimum TOTAL surface area. What's the intuition behind this? $\endgroup$ – pirsquare Dec 4 '14 at 15:36
  • $\begingroup$ If you plot the total area as a function of $r$ you get a U-shaped curve: very large when $r$ is near zero (because the curved area gets huge), very large as $r$ becomes very large (because the flat area gets huge), and not quite so large in the middle. The U-shaped curve has a "bottom" at a finite value of $r$ and it is possible to find it. Neither the flat area nor the curved area is a U-shaped function of $r$; either of these can be brought as close as you want to zero but neither one can ever be zero. $\endgroup$ – David K Dec 5 '14 at 1:42
  • $\begingroup$ Many thanks! Now it's clear. $\endgroup$ – pirsquare Dec 5 '14 at 13:35
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No, it leads to monotonous behavior.

With a single independent variable a part of the quantity should increase and a part should decrease, then only an extremum exists,verified by vanishing differential coefficient.

With two independent variables also, there should be opposite directions of variation with $x$ and $y$ in combined effect.This is verified by partial derivative with each variable.

$ A = 2 \pi r h , V = \pi r^2 h $

If the Object function A(x,y) and Constraint function V(x,y) are given, for an extremum to exist we consider with Lagrage multiplier $ -\lambda$ function to be extremized:

$ A - \lambda V $

$ \dfrac{A_x}{V_x} = \dfrac{A_y}{V_y} =\lambda $

The calculation yields $ r = r/2 $, not solvable for $r$ or $ h$,

However if Area A is considered with top and bottom areas as $ A =2 \pi r h +2 \pi r^2, $

$r$ and $h$ can be found out.

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  • $\begingroup$ Which means in the case of Total Surface Area(considered with top and bottom areas) it is not a monotonous behaviour and that is why an extremum exists $\endgroup$ – pirsquare Dec 4 '14 at 16:01
  • $\begingroup$ Yes, in that case the above calculates to $ h =r $. $\endgroup$ – Narasimham Dec 4 '14 at 17:04
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No. let's take height "h" and radius "r". $S=2\pi rh$, $V=\pi r^2h=\pi r^2(S/(2\pi r))=Sr/2$. Do you see it?

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  • $\begingroup$ Think I am missing your point. Can you explain a bit more? $\endgroup$ – pirsquare Dec 4 '14 at 15:51

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