Surface area and volume?

**maltaman** · June 30th 2009, 11:37 PM

How do you find the dimensions for cylinder that, given an area, has the least volume but the most surface area? Is there a plugin formula? Could you give me, say, an example such as the volume = 1L? (The 1 litre is not the actual question but it would help me to understand)

Thank you in advance - any help is much appreciated.

Mike

**HallsofIvy** · July 1st 2009, 03:10 AM

The only "plug-in" formulas I can think of are the formulas for volume and surface area of a cylinder: $V= \pi r^2h$ and $A= 2\pi(r^2+ rh)$ .

No, I cannot give you "an example such as the volume = 1L" because in your question, it is the surface area that is given, not the volume. I can show how to get the minimum volume given that the surface area is $1 m^2$ .

With $A= 2\pi (r^2+ rh)= 1$ , we can solve for h: $h= \frac{1- 2\pi r^2}{2\pi r}$ . Putting that into the formula for volume, $V= \frac{1}{2}(r- 2\pi r^3)$ . The extreme (maximum or minimum) values occur where the dervative is 0: $\frac{1}{2}(1- 6\pi r^2)= 0$ or $r= \sqrt{\frac{1}{6\pi}}$

**Grandad** · July 1st 2009, 09:42 AM

Hello everyone -

I'm not sure whether we're understanding the question correctly here. For a given surface area the minimum volume of a cylinder is (essentially) zero. Think of a cylinder with (almost) zero height, whose circular ends are each one-half of the given surface area, and you'll see what I mean.

Similarly, for a given volume, the surface area can be made as large as you like, by taking a small enough height.

Grandad

**maltaman** · July 1st 2009, 11:27 PM

I'm so sorry everyone, I made a mistake in the question. It was meant to be:

How do you find the dimensions for cylinder that, given an volume, has the optimum surface area? Is there a plugin formula? Could you give me, say, an example such as the volume = 1L? (The 1 litre is not the actual question but it would help me to understand)

Thank you in advance - any help is much appreciated.

Mike

**simplependulum** · July 2nd 2009, 12:02 AM

Dimensions of the cylinder : $\pi R^2 h$

Total surface area = $2 \pi R h + 2 \pi R^2$

We can let $V = \pi R^2 h ~~$ and change the subject

$h = \frac{V}{ \pi R^2 }$

Then combine the equations ,

Total surface area = $2 \pi R (\frac{V}{ \pi R^2 }) + 2 \pi R^2$

$= 2 ( \frac{V} R + \pi R^2 )$

You can see that it is a function of only R , you can find the extrema by taking its derivative .

**Tesla23** · July 2nd 2009, 09:58 PM

Here is a non calculus solution:

$ V = \pi r^2 h $

$ SA = 2 \pi r^2 + 2 \pi r h $

let $ \alpha = \frac{h}{r} $

$ SA = \frac{2 V}{h} + \frac{V}{2r} $

$ SA = 2 \pi \left(\frac{V}{\pi \alpha}\right)^\frac{2}{3} + 2 V \left(\frac{V}{\pi \alpha}\right)^\frac{-1}{3} $

$ SA = 2 \pi^\frac{1}{3} V^\frac{2}{3}\left(\frac{1}{\alpha^\frac{2}{3}} + \frac{\alpha^\frac{1}{3}}{2} + \frac{\alpha^\frac{1}{3}}{2} \right) $

using AM-GM inequality

$ SA \geq 2 \pi^\frac{1}{3} V^\frac{2}{3} 3\sqrt[3]{\frac{1}{4}} $

with equality when

$ \alpha = 2 $

so the minimum surface area occurs when the height is twice the radius. You can also find the minimum using calculus if you like.

**Grandad** · July 2nd 2009, 10:44 PM

Hello Tesla23

Welcome to Math Help Forum!

Originally Posted by Tesla23

Here is a non calculus solution:

$ V = \pi r^2 h $

$ SA = 2 \pi r^2 + 2 \pi r h $

let $ \alpha = \frac{h}{r} $

$ SA = \frac{2 V}{h} + \frac{V}{2r} $

$\color{red} SA = 2 \pi \left(\frac{V}{\pi \alpha}\right)^\frac{2}{3} + 2 V \left(\frac{V}{\pi \alpha}\right)^\frac{-1}{3} $

$ SA = 2 \pi^\frac{1}{3} V^\frac{2}{3}\left(\frac{1}{\alpha^\frac{2}{3}} + \frac{\alpha^\frac{1}{3}}{2} + \frac{\alpha^\frac{1}{3}}{2} \right) $

using AM-GM inequality

$ SA \geq 2 \pi^\frac{1}{3} V^\frac{2}{3} 3\sqrt[3]{\frac{1}{4}} $

with equality when

$ \alpha = 2 $

so the minimum surface area occurs when the height is twice the radius. You can also find the minimum using calculus if you like.

An ingenious solution!

I didn't find it easy to see how you arrived at the line I've coloured red. May I suggest:

$ V = \pi r^2 h $

$ SA = 2 \pi r^2 + 2 \pi r h $

let $ h =\alpha r $

Then $V = \pi r^3\alpha \Rightarrow r = \left(\frac{V}{\pi\alpha}\right)^{\tfrac13}$ and therefore $h = \alpha\left(\frac{V}{\pi\alpha}\right)^{\tfrac13}$

$\Rightarrow SA = 2\pi\left(\frac{V}{\pi \alpha}\right)^{\tfrac23} + 2\pi\left(\frac{V}{\pi\alpha}\right)^{\tfrac13}\al pha\left(\frac{V}{\pi\alpha}\right)^{\tfrac13}$

$= 2 \pi \left(\frac{V}{\pi \alpha}\right)^{\tfrac23} + 2 V \left(\frac{V}{\pi \alpha}\right)^{-\tfrac13}$

...etc

Grandad

**billym** · October 22nd 2009, 02:29 PM

Can I just resurrect this delightful thread?

I imagine this is a standard question, but google is so glutted with hyper-physics and stuff these days, its hard to find a standard solution.

I have a liter of water and i want to minimize the surface area of a closed cylinder.

I assume I want to minimize: $S=2\pi rh+2\pi r^2$

with the constraint $V=\pi r^2h=1$

Can somebody start me off if I wanted to use the Lagrange method?

Can I transform S and V so that

$S=f(x,y,z)=$ ...

and

$V=g(x,y,z)=$ ...

Or do I do something else?

**rn443** · October 22nd 2009, 03:14 PM

Originally Posted by billym

Can I just resurrect this delightful thread?

I imagine this is a standard question, but google is so glutted with hyper-physics and stuff these days, its hard to find a standard solution.

I have a liter of water and i want to minimize the surface area of a closed cylinder.

I assume I want to minimize: $S=2\pi rh+2\pi r^2$

with the constraint $V=\pi r^2h=1$

Can somebody start me off if I wanted to use the Lagrange method?

Can I transform S and V so that

$S=f(x,y,z)=$ ...

and

$V=g(x,y,z)=$ ...

Or do I do something else?

Set f(r, h) = 2pi*rh + 2pi*r^2 and g(r, h) = pi*r^2*h. Then you want to solve grad f = lambda * grad g, g(r, h) = 1.

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