Calc optimization

Next, we minimize. We set the first derivate equal to 0 and obtain
[eqn]
0 = l_1\left(\frac{1}{2\pi} + \frac{\sqrt{3}}{18}\right) - \frac{10\sqrt{3}}{3}
= l_1\frac{18 + 2\pi\sqrt{3}}{36\pi} - \frac{10\sqrt{3}}{3}\\
\Rightarrow l_1^* = \frac{120\pi\sqrt{3}}{18 + 2\pi\sqrt{3}}
[/eqn]

This value is approximately 22.607, which is on our interval; i.e. it is between the values 0 and 60. Indeed, you can confirm that plugging in the value to [math]A(l_1)[/math] will yield a value lower than [math]A(0)[/math] and [math]A(60)[/math].

We cut the wire into two pieces, say the first of which has length [math]l_1^*[/math]. We use that first piece to form the circle, and the second piece to to form the equilateral triangle.