Minimum/Maximum problem

[Karol]

Homework Statement

Homework Equations

Maxima/minima are where the first derivative is 0

The Attempt at a Solution

Surface area of the whole metal for creating the can (the end pieces i take as squares):
$$A=2\pi h+8r^2,~~V=\pi r^2 h~~\rightarrow~h=\frac{V}{\pi r^2}$$
$$A=2\pi \frac{V}{\pi r^2}+8r^2,~~A'=2V(-2)r^{-3}+16r=16r-\frac{4V}{r^3}$$
$$A'=0~\rightarrow~4r^4=V$$
I can't get rid of the volume V

 

[Orodruin]

You need to think more about your expression for the area contribution from the mantle area. It has the wrong physical dimension.

 

[Karol]

$$A=2\pi r h+8r^2=2\pi r \frac{V}{\pi r^2}+8r^2=2\frac{V}{r}+8r^2$$
$$A'=-2Vr^{-2}+16r,~~A'=0~\rightarrow~8r^3=V$$
No good

 

[Orodruin]

Note that you are looking for the ratio $\xi =h/2r$ that minimises the area for a fixed volume.

 

[Karol]

$$\frac{h}{2r}=\frac{\frac{V}{\pi r^2}}{\sqrt[3]{V}}$$
Can't get rid of V

 

[Orodruin]

Yes you can. You are not using the information that you have regarding how V is expressed in terms of the ratio.

 

[Karol]

$V=2\pi r^2h~$ is the only relation regarding V, so i can't express V in terms of the ratio $~\frac{h}{2r}$
besides, i used already $~V=2\pi r^2h~$ in $~A=2\pi r h+8r^2~$, so i can't use it again

 

[Orodruin]

$V=2\pi r^2h~$ is the only relation regarding V, so i can't express V in terms of the ratio $~\frac{h}{2r}$
besides, i used already $~V=2\pi r^2h~$ in $~A=2\pi r h+8r^2~$, so i can't use it again

This is an assumption you make and that assumption is wrong. Why wouldn't you be able to use it again?

 

[Karol]

$$V=2\pi r^2h,~\xi=\frac{h}{2r},~~V=2\pi r^3\xi$$
$$\xi=\frac{V^{2/3}}{\pi r^2}=\frac{(2\pi r^3\xi)^{2/3}}{\pi r^2}=\frac{\sqrt[3]{4}}{\pi^3}\xi^{2/3}$$
$$\Rightarrow~\xi^{1/3}=\frac{\sqrt[3]{4}}{\pi^3}~~\rightarrow~\xi=\frac{4}{\pi^6}$$
The answer should be $~\xi=\frac{4}{\pi}$

 

[Orodruin]

Your arithmetic is off. The $\pi^3$ is a $\pi^{1/3}$...

 

[Karol]

$$V=2\pi r^2h,~\xi=\frac{h}{2r},~~V=4\pi r^3\xi$$
$$\xi=\frac{V^{2/3}}{\pi r^2}=\frac{(4\pi r^3\xi)^{2/3}}{\pi r^2}=\frac{\sqrt[3]{4}}{\sqrt[3]{\pi}}\xi^{2/3}$$
$$\Rightarrow~\xi^{1/3}=\frac{\sqrt[3]{4}}{\sqrt[3]{\pi}}~~\rightarrow~\xi=\frac{4}{\pi}$$
Thank you Orodruin