View Full Version : Proving smallest surface area of cylinder is when h=2r

sanddrag

03-13-2005, 07:45 PM

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.

Joshua May

03-13-2005, 07:50 PM

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 havign trouble. Can someone please help. Thanks.

I think you've got it so far.

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.

sanddrag

03-13-2005, 08:02 PM

I think you've got it so far.

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.

I thought that's what I do, but I'm not sure how to solve it. Please help, final is tomorrow morning! :ahh:

Here's what I get for the derivative:

A'=4(pi)(r) +[(2r-2v)/(r^2)]

Is this right? If so, how do I solve it.

Manoel

03-13-2005, 08:04 PM

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! :ahh: ). To prove it's a minimum point, take the second partial derivative - you get a positive constant, thus, it is indeed a minimum point. :)

Kevin Sevcik

03-13-2005, 08:05 PM

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.

Kevin Sevcik

03-13-2005, 08:11 PM

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! :ahh: ). To prove it's a minimum point, take the second partial derivative - you get a positive constant, thus, it is indeed a minimum point. :)

if you try deriving just the second equation, then h is a variable, and you end up with a dh/dr term that you need a value for. substituting from the volume equation gets rid of this term because V is a constant in this case.

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

Greg Marra

03-13-2005, 08:17 PM

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...

Kevin Sevcik

03-13-2005, 08:21 PM

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...

we're talking about for a fixed volume. so you can't vary things at will.

Manoel

03-13-2005, 08:32 PM

if you try deriving just the second equation, then h is a variable, and you end up with a dh/dr term that you need a value for. substituting from the volume equation gets rid of this term because V is a constant in this case.

Yes, but I was taking the partial derivative, so you just consider h a constant to take the derivative (that's far from a formal mathematical statement, but I don't know the exact English terms to make it more appropriate :)).

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.

sanddrag

03-13-2005, 08:33 PM

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^2I can follow you here. The key IS to think of V as something like some random number (has a derivative of 0). Then when I set this equal to zero and solve I get 2V=4*pi*r^3. Then I substitue in pi*r^2*h for V and voila! it reduces down to 2h=r.

:) :) :) :)

Thanks so much everyone!

Kevin Sevcik

03-13-2005, 08:38 PM

Yes, but I was taking the partial derivative, so you just consider h a constant to take the derivative (that's far from a formal mathematical statement, but I don't know the exact English terms to make it more appropriate :)).

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.

The whole point of the problem is that h and r are varying, so you can't use a partial derivative and treat h as constant. At any rate, glad to hear you got it sanddrag. Looks like I forgot about setting A = 0 which makes it much easier to solve than I was thinking. oops.

vBulletin® v3.6.4, Copyright ©2000-2016, Jelsoft Enterprises Ltd.