# Thread: Minimum area of cylinder

1. ## Minimum area of cylinder

I have to find the minimum surface area of a right cylinder using numerical method
I have did it with differentation, I have looked back at some of my work and in all my books,
I can get that the surface area = 2pi(r^2) + 2V/r

And found that using arithmetic mean-geometric mean inequality

SA= 2pi(r^2) + 2V/r

SA= 2pi(r^2) + V/r + V/r

This the part I dont quite understand, how it changes to this
SA> 3 x cubedrt2pi{(r^2) (V/r)(V/r)}

I can just do it and be done but I realy prefer to understand what I am doing
any advise would be great thx

2. Originally Posted by pmh118
I have to find the minimum surface area of a right cylinder using numerical method
I have did it with differentation, I have looked back at some of my work and in all my books,
I can get that the surface area = 2pi(r^2) + 2V/r

And found that using arithmetic mean-geometric mean inequality

SA= 2pi(r^2) + 2V/r

SA= 2pi(r^2) + V/r + V/r

This the part I dont quite understand, how it changes to this
SA> 3 x cubedrt2pi{(r^2) (V/r)(V/r)}

I can just do it and be done but I realy prefer to understand what I am doing
any advise would be great thx
1. I assume that the volume of the cylinder is a constant.

2. The surface area of a cylinder is:

$a_s = 2 \cdot \pi r^2 + 2 \pi r \cdot h$

3. The volume of the cylinder is:

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

If you calculate $\dfrac{2V}r=2 \dfrac{\pi r^2 \cdot h}r = 2 \pi r \cdot h$ you'll get indeed the curved surface of the cylinder twice.

4. To get the minimum surface area you have to differentiate

$a_s(r)=2 \cdot \pi r^2+\dfrac{2V}r$

$a'_s(r)=4 \cdot \pi r - \dfrac{2V}{r^2}$

5. Now solve $a'_s(r)=0$ for r:

$4 \cdot \pi r - \dfrac{2V}{r^2}=0~\implies~4 \cdot \pi r^3 =2V$

So after moving some stuff around you'll get: $r = \sqrt[3]{\dfrac V{2\pi}}$

6. Now plug in this value into the equation of the surface area and you'll get your result.

3. Thank you for your response maybe I need to clarify a bit better
Yes the Volume is constant
I have already sucssesfully found the minimum Radius value with differentiation and proved it was minimum with the second derivitive
And also worked out the Surface area with this Radius value given me the minimum surface area.
But I found a new formula that works for the minimum surface area with using only the volume I just dont understand how derive this new formula

My problem is I dont understand how to get from:

2\pi}{r^2}+\frac Vr + \frac Vr

to

3\sqrt[3]{2\pi}{V^2}

I hope this LaTex works

### math geometry minimum

Click on a term to search for related topics.