Maximize and Minimize a 12" piece of wire into a square and circle 
A wire of length 12" can be bent into a circle, a square or cut into 2 pieces and make both a circle and a square. How much wire should be used for the circle if the total area enclosed by the figure(s) is to be:
a) a Maximum
b) a Minimum

What I've got so far is that the formula for the square is $A_s=\frac{1}{16}s^2$ and the circumfrance of the circle to be $P=12-c$ and area to be $A_c = \pi(\frac{P}{2\pi})^2$ where $c$ is the length of the wire for the circle and $s$ is the length of the wire for the square.
Now I know I need to differentiate these formulas to then find the max and min they both can be, but what am I differentiating with respect to? The missing variable in each of the formulas?
Also, once, I find the derivitives, what would my next steps be to minimizing and maximizing these?
And did I set the problem up correctly?
Thanks for any help

Let $s$ be the circumference of the square. Then the circumference of the circle is $12-s$ (because that's what is left from the wire). Now you already computed the formulas $A_{\mathrm{square}}(s) = \frac{1}{16} s^2$ and $A_{\mathrm{circle}}(s) = \frac{1}{4\pi}(12 - s)^2$. The total area is $A(s) = A_{\mathrm{square}}(s) + A_{\mathrm{circle}}(s)$, where $s \in [0,12]$ is the variable. To find the extrema (maximum/minimum) of this function, a necessary condition is $A'(s) = 0$ (differentiate with respect to $s$) when $0 \lt s \lt 12$ and you need also consider $A(0)$ and $A(12)$.
So the task you need to do is to differentiate $A(s)$ with respect to $s$, solve $A'(s) = 0$ for $s$ (there will be only one solution $s_0$). Now the maximum among $A(0)$, $A(12)$ and $A(s_0)$ will be the maximum and the minimum among them will be the minimum of $A(s)$. It may also help if you sketch the graph to convince yourself of the solution.

Here's a small sanity check: The circle is the geometric figure that encloses the largest area among all figures with the same circumference, so the maximum should be achieved for $s = 0$. Since enclosing two figures needs more wire than enclosing a single one, the minimum should be achieved at $s_0$.

Added:
Since the results you mention are a bit off, let me show you what I get:
First $$A(s) = \frac{1}{16}s^2 + \frac{1}{4\pi}(12-s)^2.$$
Differentiating this with respect to $s$ I get
$$A'(s) = \frac{1}{8}s - \frac{1}{2\pi}(12-s)$$
Now solve $A'(s) = 0$ to find $$s_0 = \frac{12}{1+\frac{\pi}{4}} \approx 6.72$$
Plugging this in gives me $A(s_0) \approx 5.04$. (No warranty, I hope I haven't goofed)
