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A wire 5 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each.

I understand most of this but I'm having trouble finding the right terms for the triangle's area and optimization. Any help would be appreciated.

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2 Answers 2

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Consider $x=$ the length of the part used to make equilateral triangle.

Then each side of the triangle $= x/3$ and area of the triangle $= \frac{\surd{3}x^2}{36}$.

Area of the circle $=\pi(\frac{5-x}{2\pi})^2$

Let $f(x)=\frac{\surd{3}x^2}{36}+\pi(\frac{5-x}{2\pi})^2$

Let $f'(x)=\frac{\surd{3}x}{18}-\frac{5-x}{2\pi}=0$, then find $x$.

Also $f''(x)=\frac{\surd{3}}{18}+\frac{1}{2\pi}>0$

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Let $x$ an $5-x$ be the length of wires and bent into an equilateral triangle and a circle respectively. Area of equilateral triangle given side length $\dfrac{x}{3}$ is $\dfrac{\sqrt{3}}{4}\left(\dfrac{x^2}{9}\right)$ and that of circle given circumference $5-x$ is $\dfrac{(5-x)^2}{4\pi}$. Total area $f(x)=\dfrac{\sqrt{3}}{4}\left(\dfrac{x^2}{9}\right)+\dfrac{(5-x)^2}{4\pi}$. To minimize we differentiate, find the critical points and determine the local minima. $f'(x)=\dfrac{\sqrt{3}}{18}x+\dfrac{(x-5)}{2\pi}=0$ we get $x=\dfrac{45}{9+\pi\sqrt{3}}$. By second derivative test we see minima occurs at $x=\dfrac{45}{9+\pi\sqrt{3}}\approx3.11604$

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