<s>[INST] 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 $pi$."
 [/INST] 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 $\displaystyle (10-x) = 10-\frac{10\pi}{\pi+4} = \frac{40}{\pi+4}$
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