# Calculus Optimization Problem: Wire Triangle and Circle

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.

## 2 Answers

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

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