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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.

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Let $L$ be the length of the wire, $x, y$ be the side of the square and the radius of the circle, $S$ be the sum of the areas of the circle and the square. Then: $S = x^2 + \pi y^2, 4x+2\pi y = L$. By the Cauchy-Schwarz inequality: $L^2 = (4x+2\sqrt{\pi}\sqrt{\pi}y)^2\leq (4^2+4\pi)(x^2+\pi y^2)\Rightarrow S \geq \dfrac{L^2}{16+4\pi}$. Equality occurs when $\dfrac{4}{x}=\dfrac{2\sqrt{\pi}}{\sqrt{\pi} y}=\dfrac{2}{y}\to x = 2y\to r = \dfrac{2\pi y}{4x}=\dfrac{\pi}{4}$

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  • $\begingroup$ Thank you very much I liked this approach more then mine also. I got the same by taking the derivative with the other hint which I'm not sure why I forgot that. $\endgroup$ – HighSchool15 Aug 2 '15 at 4:30
  • $\begingroup$ You are welcome ! $\endgroup$ – DeepSea Aug 2 '15 at 4:32
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Differentiate with respect to x the set the dA/dx to 0 and solve for x. dA/dx= -2(y-x)/4pie +2x/16 0 = -2(y-x)/4pie + 2x/16 x=4y/pie+4

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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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