<s>[INST] Minimize the area of a wire divided into a circle and square. 
A wire is divided into two parts. One part is shaped into a square, and the other part is shaped into a circle. Let r be the ratio of the circumference of the circle to the perimeter of the square when the sum of the areas of the square and circle is minimized. Find r.

My attempt so far 
The total length of the wire is $y$ the length used for the square is $x$ therefore the perimeter of the square is $x$ the area of the square is then $\dfrac{x^2}{16}$, the length for the circle is $y-x$ and the circumference is therefore $y-x$ 
$$R=\dfrac{y-x}{2\pi}$$
Area of the circle is then $$\dfrac{(y-x)^2}{4\pi}$$
I need to minimize $$\dfrac{x^2}{16} + \dfrac{(y-x)^2}{4\pi}$$
In order to solve for $$r=\dfrac{y-x}{x}$$ 
Now I'm a little stuck. Thanks for any help. 
 [/INST] For Minimization of $$\displaystyle \frac{x^2}{16}+\frac{(y-x)^2}{4\pi}\;,$$ We Also use $\bf{Cauchy-Schwartz}$ Modified Inequality.
Which is $$\displaystyle \frac{A^2}{X}+\frac{B^2}{Y}\geq\frac{(A+B)^2}{X+Y}$$ and equality hold When $\displaystyle \frac{A}{X} = \frac{B}{Y}$.
So $$\displaystyle \frac{x^2}{16}+\frac{(y-x)^2}{4\pi}\geq \frac{(x+y-x)^2}{16+4\pi}=\frac{y^2}{16+4\pi}.$$ and equality hold when $\displaystyle \frac{x}{16}=\frac{y-x}{4\pi}$
So we get $$\displaystyle \displaystyle y-x = \frac{4\pi \cdot x}{16}$$ . So we get  $$\displaystyle r = \frac{y-x}{x} = \frac{4\pi}{16} = \frac{\pi}{4}$$
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