Optimization and maximum geometry What is the side length of the largest square that will fit inside an equilateral triangle with sides of length 1.
I created two equations:
Square: Area=$x^2$
Triangle: Area= $\sqrt{3 /4}$
However, how can I find the maximum??
I did $\sqrt{3 /4}=x^2$ ang got fourth root $3 / 2$ which is wrong


$1)$ Suppose that $DEFG$ is a square. Let's call $x$ the side of the square.
We also have that $\Delta ADE$ is an equilateral triangle, so $AD=x$. 
Now, for the triangle $BDG$ we have:
$$\sin 60°=\frac{DG}{BD}=\frac{x}{BD} \rightarrow BD=\frac{2\sqrt{3}x}{3}$$
$$AB=1=AD+BD=x+\frac{2\sqrt{3}x}{3} \rightarrow x=\frac{3}{3+2\sqrt{3}}=2\sqrt{3}-3$$
It means that doesn't make sense talk about largest square because we have just one and its side is $x=2\sqrt{3}-3$.
$2)$ What could make sense is talk about the biggest rectangle inside the triangle. In order to do that we can call $DE=x$ and then $AD=x$. Let's also call $DG=y$, so
$$\sin 60°=\frac{DG}{BD}=\frac{y}{BD} \rightarrow BD=\frac{2\sqrt{3}y}{3}$$
and then
$$AB=1=AD+BD=x+\frac{2\sqrt{3}y}{3}$$
and we have to maximize:
$$A=xy=\left(1-\frac{2\sqrt{3}y}{3}\right)y$$
and the maximum happens for (using the quadratic standard approach)
$$y=\frac{\sqrt{3}}{4} \rightarrow x=\frac{1}{2}$$
what give us 
$$A_{max}=\frac{\sqrt{3}}{8}$$
