Tin can with the least amount of tin

[Elpinetos]

Homework Statement

Find the smallest r and h for a tin can with Volume a) 1L b) V so that the amount of tin used is minimal

Homework Equations

$V=r^{2}\pi h$
$A=2\pi(r^{2}+h)$

The Attempt at a Solution

Since the area has to be minimal, I expressed r from V
$r^{2}=\frac{V}{\pi h}$
and inputed it into the equation of A
$A=2\pi(\frac{V}{\pi h}+h)$

Now I get the first derivative:
$\frac{dh}{dA}=2\pi(\frac{-V}{\pi^{2}h^{2}}+1)$

I set this to equal 0, do the algebra and end up with
$-1=\frac{-V}{\pi^{2}h^{2}} \\<br /> -\pi^{2}h^{2}=-V \\<br /> \pi^{2}h^{2}=V \\ <br /> h^{2}=\frac{V}{\pi^{2}}\\<br /> h=\frac{\sqrt{V}}{\pi}$

When I input this into the equation for r I get:
$r=\frac{\sqrt[4]{V}}{\sqrt{\pi}}$

If I set $V=1$ it gives me ~0.32 for h and ~0.56 for r

The solution book tells me the solution is
$r=\sqrt[3]{V(2\pi)}$
$h=2\sqrt[3]{V(2\pi)}$

Which would give me totally different values for h and r

Where did I make a mistake? :(

 

[Mark44]

I cleaned up your LaTeX, which was pretty hard to read.

Homework Statement

Find the smallest r and h for a tin can with Volume a) 1L b) V so that the amount of tin used is minimal

Homework Equations

$V=r^{2}\pi h$
$A=2(r^{2}+h)$

Your formula for area is wrong. The total surface area should include the cylinder, which is made from a rectangular piece, plus two circular end pieces.

The Attempt at a Solution

Since the area has to be minimal, I expressed r from V
$r^{2}=\frac{V}{\pi h}$
and inputed it into the equation of A
$A=2\pi(\frac{V}{\pi h}+h)$

Now I get the first derivative:
$\frac{dh}{dA}=2\pi(\frac{-V}{\pi^{2}h^{2}}+1)$

I set this to equal 0, do the algebra and end up with
$-1=\frac{-V}{\pi^{2}h^{2}} \\<br /> -\pi^{2}h^{2}=-V \\<br /> \pi^{2}h^{2}=V \\ <br /> h^{2}=\frac{V}{\pi^{2}}\\<br /> h=\frac{\sqrt{V}}{\pi}$

When I input this into the equation for r I get:
$r=\frac{\sqrt[4]{V}}{\sqrt{\pi}}$

If I set $V=1$ it gives me ~0.32 for h and ~0.56 for r

The solution book tells me the solution is
$r=\sqrt[3]{V(2\pi)}$
$h=2\sqrt[3]{V(2\pi)}$

Which would give me totally different values for h and r

Where did I make a mistake? :(

 

[Elpinetos]

Thank you, but I looked into my formula booklet just now and it tells me that the area is $2r^{2}\pi+2r\pi h$which is $2\pi(r^{2}+h)$

I just saw that I forgot to write it into my LaTeX above, but I still calculated with it, so that can't be it Oo

EDIT: I just saw I forgot the second r...
Nevermind... Thank you

>.<