# A square and a equilateral triangle are to be formed out of the same piece of wire. The wire is 6 inches long. How do you maximize the total area the square and the triangle contain?

May 13, 2016

$\frac{4 \sqrt{3} L}{9 + 4 \sqrt{3}}$ for the square
$\frac{9 L}{9 + 4 \sqrt{3}}$ for the equilateral triangle
Here $L = 6$

#### Explanation:

Let be $L = s + t$ the total length as the addition of $s$ the length used by the square and $t$ the length used by the triangle.
The square area is ${a}_{s} = {\left(\frac{s}{4}\right)}^{2}$ and the equilateral triangle area is given by ${a}_{t} = \left(\frac{t}{6}\right) \sqrt{{\left(\frac{t}{3}\right)}^{2} - {\left(\frac{t}{6}\right)}^{2}} = {t}^{2} / \left(12 \sqrt{3}\right)$
the total area is then $a = {a}_{s} + {a}_{t} = {s}^{2} / 16 + {t}^{2} / \left(12 \sqrt{3}\right)$
but $t = L - s$ then $a = {\left(L - s\right)}^{2} / \left(12 \sqrt{3}\right) + {s}^{2} / 16$
The area critical point is determined doing $\frac{\mathrm{da}}{\mathrm{ds}} = 0$ and obtaining $s = \frac{4 \sqrt{3} L}{9 + 4 \sqrt{3}}$ and also $t = \frac{9 L}{9 + 4 \sqrt{3}}$