Minima problem? This is a question in my textbook which I can't solve. Any help would be appreciated, thanks.
"A piece of wire 10 metres long is cut into two portions. One piece is bent to form a circle, and the other piece to form a square. Find the circumference of the circle if the sum of areas of the circle and square is to be a minimum. Give your answer in terms of $pi$."

It' fairly easy to solve a more general problem: You cut the wire of length $1$ in two pieces, $ x$ and $1-x$. From one piece you form a regular $n$-gon, from the other a regular $m$-gon.  Minimize the sum of the area of both.
For any regular $n$-gon let $r_n$ the radius of its inscribed circle and $\pi_n$ the ratio of its circumference to its diameter. (We'll get $\pi_n=n\cdot\tan(\pi/n)$, so, e.g., $\pi_3=3\sqrt3$, $\pi_4=4$ and $\pi_6=2\sqrt3$ and so on.  We may show that $\pi_\infty=\pi$.)
Now it's easy to verify that the circumference $L_n$ of the $n$-gon is 
$L_n=2\cdot\pi_n\cdot r_n$ and its area is $A_n=\pi_n\cdot r_n^2$.  From $r_n=L_n/2\pi_n$ we conclude $A_n=L_n^2/4\pi_n$.
Thus the sum of the area equals
$$\frac{x^2}{4\pi_n}+\frac{(1-x)^2}{4\pi_m}.$$
By standard methods it arrives its minimum in 
$$x=\frac{\pi_n}{\pi_n+\pi_m}.$$
So cut a wire of length $r$ in two pieces of length $\frac{r\pi_n}{\pi_n+\pi_m}$ and $\frac{r\pi_m}{\pi_n+\pi_m}$.  The the minimal area is
$$\left(\frac{r}{2}\right)^2\frac{1}{\pi_n+\pi_m}.$$
For $n=3$ and $m=6$ we'll get $0.6r$ and $0.4r$ with minimal area $(r/2)^2\cdot\frac{1}{5\sqrt3}$.
