# Given a fixed volume for a solid cylinder, is it possible to find the minimum or maximum curved surface area?

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?

• No. Why do you ask? – David K Dec 4 '14 at 13:48
• 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? – pirsquare Dec 4 '14 at 15:36
• 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. – David K Dec 5 '14 at 1:42
• Many thanks! Now it's clear. – pirsquare Dec 5 '14 at 13:35

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.

• 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 – pirsquare Dec 4 '14 at 16:01
• Yes, in that case the above calculates to $h =r$. – Narasimham Dec 4 '14 at 17:04

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?

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