# Thread: Surface area and volume?

1. ## 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

2. 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 $r= \sqrt{\frac{1}{6\pi}}$

3. ## 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.

4. 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

5. 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 .

6. 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.

7. 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

8. 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?

9. 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.