# Thread: Minimizing cylinder material - find the ratio of h to r

1. ## Minimizing cylinder material - find the ratio of h to r

I would appreciate help on this question:In constructing a cylindrical can of given volume, nothing is wasted in making the side of the can. But the top and bottom are cut from square sheets and the remainder are wasted. Find the ratio of h to r so that the material used is a minimum.

2. ## Re: Minimizing cylinder material - find the ratio of h to r

As I understand your question, you will want the radius of the circle (top and bottom) to be half the height of the square so that the circle is perfectly encased in the square, thus wasting the lowest amount of material.

$\displaystyle 2r=h\\\\2=\frac{h}{r}$

The ratio is therefore $\displaystyle \frac{1}{2}$

3. ## Re: Minimizing cylinder material - find the ratio of h to r

Given a cylinder with radius r and volume V, can you write an expression for how much material M is used? You would then find the minimum of M as a function of r.

- Hollywood

4. ## Re: Minimizing cylinder material - find the ratio of h to r

Are you assuming that the top and bottom are cut from the same size square as is used to make the sides? Paze appears to be assuming that.

5. ## Re: Minimizing cylinder material - find the ratio of h to r

Originally Posted by HallsofIvy
Are you assuming that the top and bottom are cut from the same size square as is used to make the sides? Paze appears to be assuming that.
Nono. If we look at his cylinder from the top down, it will look like this (the material being the square): http://s23.postimg.org/3wkg8ho23/cylinder.png

h in my equation represents the height of the square, not the cylinder.

Obviously this is not calculus as the post's section suggests, but this is how I understand OP's problem.

6. ## Re: Minimizing cylinder material - find the ratio of h to r

Here is my understanding of the problem:

The sides are cut without waste, so an area of $\displaystyle 2\pi{r}h$ is used.
The top and bottom are circular, and are cut out of squares, so instead of an area of $\displaystyle \pi{r^2}$ times 2 for top and bottom, an area of $\displaystyle 4r^2$ (times 2) is used.

The cylinder has volume $\displaystyle V=\pi{r^2}h$.

So the total area is $\displaystyle A=2\pi{r}h + 8r^2$. Substituting $\displaystyle h=\frac{V}{\pi{r^2}}$ gives $\displaystyle A=\frac{2V}{r}+ 8r^2$.

The volume V is fixed - we want to minimize A by varying r. So $\displaystyle 0=\frac{dA}{dr}=-\frac{2V}{r^2}+16r$, $\displaystyle r^3=\frac{V}{8}$.

The problem asked for the ratio of h to r: $\displaystyle \frac{h}{r}=\frac{V}{\pi{r^3}}=\frac{8}{\pi}$.

- Hollywood