# Surface area and volume?

• Jun 30th 2009, 11:37 PM
maltaman
Surface area and volume?
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
• Jul 1st 2009, 03:10 AM
HallsofIvy
The only "plug-in" formulas I can think of are the formulas for volume and surface area of a cylinder: $\displaystyle V= \pi r^2h$ and $\displaystyle A= 2\pi(r^2+ rh)$.

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 $\displaystyle 1 m^2$.

With $\displaystyle A= 2\pi (r^2+ rh)= 1$, we can solve for h: $\displaystyle h= \frac{1- 2\pi r^2}{2\pi r}$. Putting that into the formula for volume, $\displaystyle V= \frac{1}{2}(r- 2\pi r^3)$. The extreme (maximum or minimum) values occur where the dervative is 0: $\displaystyle \frac{1}{2}(1- 6\pi r^2)= 0$ or $\displaystyle r= \sqrt{\frac{1}{6\pi}}$
• Jul 1st 2009, 09:42 AM
Surface area and volume of a cylinder
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.

• Jul 1st 2009, 11:27 PM
maltaman
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
• Jul 2nd 2009, 12:02 AM
simplependulum
Dimensions of the cylinder : $\displaystyle \pi R^2 h$

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

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

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

Then combine the equations ,

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

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

You can see that it is a function of only R , you can find the extrema by taking its derivative .
• Jul 2nd 2009, 09:58 PM
Tesla23
Here is a non calculus solution:

$\displaystyle V = \pi r^2 h$

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

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

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

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

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

using AM-GM inequality

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

with equality when

$\displaystyle \alpha = 2$

so the minimum surface area occurs when the height is twice the radius. You can also find the minimum using calculus if you like.
• Jul 2nd 2009, 10:44 PM
Hello Tesla23

Welcome to Math Help Forum!
Quote:

Originally Posted by Tesla23
Here is a non calculus solution:

$\displaystyle V = \pi r^2 h$

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

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

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

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

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

using AM-GM inequality

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

with equality when

$\displaystyle \alpha = 2$

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:

$\displaystyle V = \pi r^2 h$

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

let $\displaystyle h =\alpha r$

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

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

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

...etc

• Oct 22nd 2009, 02:29 PM
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: $\displaystyle S=2\pi rh+2\pi r^2$

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

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

Can I transform S and V so that

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

and

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

Or do I do something else?
• Oct 22nd 2009, 03:14 PM
rn443
Quote:

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: $\displaystyle S=2\pi rh+2\pi r^2$

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

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

Can I transform S and V so that

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

and

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

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.