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.
Best Answer
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 $x=\dfrac{45}{9+\pi\sqrt{3}}\approx3.11604$