Minimizing cost materials for a cylindrical can which must hold 100 cubic inches (Max-Min pre-calc) Using a calculator, find the minimum cost of materials to create a cylindrical can that must hold 100 cubic inches. The top/bottom of the can can cost(in dollars) 0.14 per square inch, while the sides can cost only 0.07 per square inch.
I'm slightly lost on how to approach the problem.
I'm thinking...
area: 100=pi(r^2)h
 A: In an attempt to answer the question with only the tools specified:
As others have explained, the cost, $C$ of the can may be written as $$C =0.07 \times 2 \pi rh+2 \times 0.14 \pi r^2 $$But $h$ and $r$ are related through the volume, $V$, $$V=\pi r^2h$$
Eliminating $h$, using $100$ for the volume $V$, and simplifying, we obtain$$C=\frac{14}{r}+0.28 \pi r^2$$
Multiplying through by $r$ and rearranging, we obtain the following cubic equation in $r$: $$0.28 \pi r^3-Cr+14=0$$For positive $C$ (as in this problem), this cubic always has a single real root for some negative $r$. In addition, the equation has either two complex roots, or two positive real roots (possibly identical), depending on the specific value of $C$.
A graphing calculator, or one with the capability of solving cubic equations, can be used to find, by trial and error, the smallest possible value of C that produces a positive root.  That positive root is the minimum cost radius for the can.
Calculus is easier... 
A: What you quoted as area is actually volume. We know that because it has three dimensions: radius, radius again because it's squared, and height. It's a good habit to check that sort of thing so you don't make silly errors later on.
But yes, that's your formula for the volume of the can. What you want to do now is find the surface area of the ends and the sides, because at the end of the day, the problem is about surface area. It wants us to minimize the equation $y = .14A_{ends} + .07A_{sides}$, where $y$ is the cost.
$A_{ends} = 2 \times \pi r^2$. (Why is it multiplied by 2?)
$A_{side} = h \times c$ where $c$ is the circumference. The side of the can would be a rectangle when you unroll it, where its height is the height of the can and the base is the circumference. If you haven't remembered by now, the circumference is $2\pi r$.
So what we're going to do is use the formula we know for volume and solve it for $h$, then plug it into out equation for area.
$100 = \pi r^2 h$, so $h = \frac{100}{\pi r^2}$. You should verify this. 
Therefore we have $A_{side} = \frac{100}{\pi r^2} \times 2\pi r = \frac{200}{r}$, and plugging into our formula for cost, we have $y = .14 \times 2\pi r^2 + .07\frac{200}{r}$. 
At this point... you actually need calculus to get a precise answer, and I see that this question is tagged as precalculus. In the context of precalculus, the best thing to do is graph your function of cost and look at the graph to see for which value of $r$ it is minimized.
A: The volume (not area) formula is $V = \pi {r^2}h = 100$. The area of the side of the can is that of a rectangle, once you make a vertical slit in the side and flatten it out. The height of the rectangle is that of the can, and the length of the rectangle is the circumference of the top or bottom circle.
Since the side and ends of the can have different costs, the overall cost should be treated as the sum of the costs of the top, bottom, and side, which is
$$C = .14 \cdot \pi {r^2} + .14 \cdot \pi {r^2} + .07 \cdot 2\pi rh$$
Use the volume equation to solve for one variable ($r$ or $h$) in terms of the other one, then substitute into the cost formula to get the cost in terms of only one variable.
You should be able to continue from there.
A: And the function, which has to minimize is:
price * (surfaces of bottom + top) + price * side
$f(r,h)=0.14 \cdot 2 \cdot \pi \cdot r^2+0,07 \cdot 2 \cdot \pi \cdot r \cdot h$
You can solve the equation $100=h \cdot \pi \cdot r^2$ for h and insert the expression of h in the function  f(r,h).
Then derive f(r) in respect to r and set the derivation equal to $0$: $f'(r)=0$
Now solve the equation and find $r^*$
greetings,
calculus
