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#1




Proving smallest surface area of cylinder is when h=2r
I need to prove that surface area A of a cylinder (of volume V) is smallest when the height, h equals twice the radius, r.
We have two equations 1. V=(pi)(r^2)(h) 2. A=2(pi)(r^2) + 2(pi)(r)(h) Here's what I think I need to do in order to prove the smallest A occurs when h=2r. I think I need to take equation 1 and solve for h which comes out to h=V/(pi)(r^2). Then I need to substitue that in for h in equation 2. This would make equation 2 become: A=2(pi)(r^2) + 2(pi)(r)[V/(pi)(r^2)] Now, I know this can be simplified, but I'm not sure if I'm doing it right. I think it would go to: A=2(pi)(r^2) + 2V/r Is this right? If so, what do I need to do now? I know I have to take the derivative to find the absolute minium. This is where I'm having trouble. Can someone please help? Thanks. Last edited by sanddrag : 03132005 at 07:48 PM. 
#2




Re: Proving smallest surface area of cylinder is when h=2r
Quote:
I believe the next step to go is to find the derivative of A=2(pi)(r^2) + 2V/r, set that to zero, solve it, and do a sign test to show that this is a minimum point. Sorry, but I haven't done optimization problems in a few months, so I might be a little rusty, but this should set you on the right track. 
#3




Re: Proving smallest surface area of cylinder is when h=2r
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Here's what I get for the derivative: A'=4(pi)(r) +[(2r2v)/(r^2)] Is this right? If so, how do I solve it. 
#4




Re: Proving smallest surface area of cylinder is when h=2r
Sanddrag,
You don't need any volume information at all. Use your second formula and take the partial derivative of A in respect to r, set it to zero and you'll achieve the desired result (except for a nasty negative sign  maybe it's irrelevant, maybe I did something wrong! ). To prove it's a minimum point, take the second partial derivative  you get a positive constant, thus, it is indeed a minimum point. 
#5




Re: Proving smallest surface area of cylinder is when h=2r
Josh has the rest of the process right. It looks like you're going to get a 3rd order polynominial when you get dA/dr. So you should get 3 points where it equals zero. After you find the minimum, you'll have an answer for R in terms of V. You'll need to plug this back into your equation for h in terms of V and r to show that h = 2r.

#6




Re: Proving smallest surface area of cylinder is when h=2r
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sanddrag: V is a constant in this case since you said yourself for a given volume treat it as another number. Thus: A' = 4*pi*r  2*V/r^2 
#7




Re: Proving smallest surface area of cylinder is when h=2r
Perhaps I'm missing something really obvious here, but...
Wouldn't the smallest surface area occur when h and r are as close to 0 as possible? I mean, are we talking about in relation to something? In finite terms it'd seem that you can get a smaller surface area when you reduce either of those numbers... 
#8




Re: Proving smallest surface area of cylinder is when h=2r
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#9




Re: Proving smallest surface area of cylinder is when h=2r
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I agree that the height and the radius are constrained by the fixed volume, but I don't see how that changes my solution. It seems Sanddrag is taking Calculus A, or I, or 101, or whatever you call it in the USA. I didn't learn about partial derivatives until Calculus II. 
#10




Re: Proving smallest surface area of cylinder is when h=2r
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Thanks so much everyone! 
#11




Re: Proving smallest surface area of cylinder is when h=2r
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