# Extremely hard minimization problem

• Oct 4th 2008, 12:47 PM
SuperTyphoon
Extremely hard minimization problem
This one is incredibly hard. I have absolutely no idea how to solve it.

"A closed cylindrical can is to have a volume of V cubic units. Show that the can of minimum surface area is achieved when the height is equal to the diameter of the base."

I've gotten as far as finding the derivative of the Area by solving for r, but i'm just stuck on this one.
• Oct 4th 2008, 01:48 PM
Chris L T521
Quote:

Originally Posted by SuperTyphoon
This one is incredibly hard. I have absolutely no idea how to solve it.

"A closed cylindrical can is to have a volume of V cubic units. Show that the can of minimum surface area is achieved when the height is equal to the diameter of the base."

I've gotten as far as finding the derivative of the Area by solving for r, but i'm just stuck on this one.

Note that we have two equations that we can set up:

$\pi r^2h=V$ ---------------[Volume of the cylinder]
$S=2\pi r^2+2\pi r h$ -------[Surface area of cylinder]

I would first solve for height in the volume equation, and then substitute that into the surface area equation:

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

Subbing in, we see that $S=2\pi r^2+2\pi r\frac{V}{\pi r^2}$

Now simply a little bit:

$S=2\pi r^2+\frac{2V}{r}$

Thus, $\frac{\,dS}{\,dr}=4\pi r-\frac{2V}{r^2}$

Setting $\frac{\,dS}{\,dr}=0$, we see that $\frac{2V}{r^2}=4\pi r\implies \frac{2V}{\pi}=4r^3\implies 2r=\sqrt[3]{\frac{4V}{\pi}}$. Note that $2r=d$, where $d$ is the diameter of the base. Thus, $\color{red}\boxed{d=\sqrt[3]{\frac{4V}{\pi}}}$

Now what is h?

Note that $\pi r^2h=V$

Thus,

${\color{red}4}\pi r^2 h={\color{red}4}V\implies \pi d^2h=4V\implies h=\frac{4V}{\pi d^2}\implies h=\frac{4V}{\pi}\cdot\frac{1}{\left[\sqrt[3]{\frac{4V}{\pi}}\right]^2}\implies$

$h=\frac{4V}{\pi}\cdot\left(\frac{4V}{\pi}\right)^{-\frac{2}{3}}\implies \color{red}\boxed{h=\sqrt[3]{\frac{4V}{\pi}}}$

We have shown that the height and diameter are the same.

Does this make sense?

--Chris
• Oct 4th 2008, 02:54 PM
SuperTyphoon
Wow! That was incredible. Thank you so much. It makes perfect sense to me! I wish i could do that i on my own...
• Oct 6th 2008, 02:11 PM
SuperTyphoon
As great is this is, i have found an arithmetic error in the solution that has left me hanging once again!

$(2V/pi)=4r^3$ He divided each side by 2 to get...

$2r^3=(4V/pi)$ But wouldn't dividing by 2 give you $2r^3=(V/pi)$ ?

You would then get $r=\sqrt[3](V/2pi)$ thus $d=2r=2\sqrt[3](V/pi)$.

What would you do now?
• Oct 6th 2008, 07:07 PM
Chris L T521
Quote:

Originally Posted by SuperTyphoon
As great is this is, i have found an arithmetic error in the solution that has left me hanging once again!

$(2V/pi)=4r^3$ He divided each side by 2 to get...

$2r^3=(4V/pi)$ But wouldn't dividing by 2 give you $2r^3=(V/pi)$ ?

You would then get $r=\sqrt[3](V/2pi)$ thus $d=2r=2\sqrt[3](V/pi)$.

What would you do now?

I didn't divide through by two! I actually multiplied through by two and then skipped some steps. May I should have included the extra steps... (Doh)

$\frac{2V}{\pi}=4r^3\implies {\color{red}2}\cdot\frac{2V}{\pi}={\color{red}2}\c dot 4r^3\implies \frac{4V}{\pi}=8r^3\implies \frac{4V}{\pi}=(2r)^3\implies 2r=\sqrt[3]{\frac{4V}{\pi}}$

Does this clarify things?

--Chris
• Oct 7th 2008, 02:33 AM
SuperTyphoon
Ahhhh yessss. That makes a big difference! Thanks for clearing that up. My calc teacher always discourages skipping steps for the same reason that happened here.

Thanks again!