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 $pi$."
 A: HINT:
If the radius of the circle is $=r$ meter, the perimeter of the circle $=2\pi r$ meter
If one side of the square is $=a$ meter, the perimeter of the sqaure $=4a$ meter
So, we have $2\pi r+4a=10=a=\frac{10-2\pi r}4$
and the sum of area $=\pi r^2+a^2=f(a)$ to be maximized
A: 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 $(r/2)^2\cdot\frac{1}{5\sqrt3}$.
A: Given Total length of wire $10$ meter, Now Let $x$ meter cut from it and form a Circle and
remaining $(10-x)$ form a Square.
So Here Radius of Square is $\displaystyle \frac{x}{2\pi}$ and Length of square is $\displaystyle \frac{(10-x)}{4}$
Now Area of Circle $\displaystyle \pi\cdot \frac{x^2}{4\pi^2} = \frac{x^2}{4\pi}$ and Area of Square $\displaystyle \frac{(10-x)^2}{16}$
So Total area $\displaystyle A = \frac{x^2}{4\pi}+\frac{(10-x)^2}{16}.$
Now here we have to minimize $A$
So using $\bf{Cauchy-Schwartz}$ Inequality, We get
$$\displaystyle A = \frac{x^2}{4\pi}+\frac{(10-x)^2}{16}\geq \frac{x+10-x}{4\pi+16}  = \frac{10}{4\pi+16}$$
and equality hold when $$\displaystyle \frac{x}{4\pi} = \frac{10-x}{16}\Rightarrow x=\frac{10\pi}{\pi+4}$$
So length of wire to form a circle $\displaystyle x = \frac{10\pi}{\pi+4}$ and 
length of wire to form a square $\displaystyle (10-x) = 10-\frac{10\pi}{\pi+4} = \frac{40}{\pi+4}$
