Calculus Question pertaining to volume I'm having a lot of trouble with this problem, so I need some help. 
A company plans to make aluminum can, each with a lid and containing a volume of 2,000 cubic centimeters.
a) Find the dimensions of the can that will minimize the amount of aluminum used. (V = πr^2h)
b) What is the relation between the height and the base diameter? Is this relation the same for a can of any volume if the surface area is minimized?
Like I said Im having a ton of trouble with this and don't even know were to start. Im teaching myself and the book really sucks. 
Thanks!
 A: Because typing $V$ is easier than typing $2000$ (and with an eye on part (b)), we write $2000$ instead of $V$.  Let us explore the consequences of choosing a base radius of $r$.  Then the height $h$ of the cylindrical can will be given by $h=\frac{V}{pi r^2}$.
We want to minimize the surface area of the can. That surface area will depend on $r$. Let us call it $S(r)$.
The surface is made up of (i) the top and bottom disks and (ii) the rest.
(i) The combined area of the top and bottom disks is $2\pi r^2$.
(ii) Imagine removing the top and bottom of the can, then making a vertical cut with tin shears, and flattening out the metal. We get a rectangle which is $2\pi r \times h$. So it has area $2\pi rh$. 
It follows that 
$$S(r)=2\pi r^2 +2\pi rh.$$
Use $h=\frac{V}{\pi r^2}$ to get $S(r)$ in terms of $r$ only. We get
$$S(r)=2\pi r^2 +\frac{2}{r}.$$
Now go through the usual calculus machinery to find the $r$ that minimizes $V(r)$.
A: You want to minimize the surface area, for a cylinder SA=$2\pi r^2 + \pi r h$. What does it mean to minimize this equation? In order to do it you will need to use your volume constraint $V=\pi r^2 h = 2000$ to solve for $r$ or $h$ in your SA equation. Part b should follow, so this should be enough.
A: Using your cylindrical volume equation, $V = \pi r^2h$, substitute the volume required to obtain a function $h(r)$ which will tell you the height needed for any radius that gives a cylinder with the correct volume.
Now use the equation of surface area $S\!A = 2(\pi r^2) + (2 \pi r) h$, and substitute the $h(r)$ you just derived to get $S\!A(r) = 2(\pi r^2) + (2 \pi r) h(r)$. This is a function of surface area in terms of radius, so to minimize it, differentiate with respect to $r$, and solve for roots:
$${\mathrm{d}S\!A \over \mathrm{d}r} =0$$
