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

Proof with some words

You can create a rectangle inside your triangle. Let $x_B$ and $x_G$ the $x$-coordinates of the blue and green points, respectively. It is clear that $y_B = y_G = 0$. Moreover, it is easy to see that:
$$x_B < 0, x_G >0 ~\text{and}~ x_B = -x_G.$$
The base of the rectangles is $$b = x_G - x_B = 2 x_G.$$
The height of the rectangle is given by the $y$-coordinate of the yellow or red point, let's call them $y_Y$ and $y_R$. Note also that the $x$-coordinates of these two points are:
$$\begin{cases}
x_R = x_B \\
x_Y = x_G
\end{cases}.
$$
Then, using the equation of the diagonal side of your triangle, we have that:
$$\begin{cases}
y_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_B \\
y_Y = \frac{\sqrt{3}}{2} - \sqrt{3}x_Y = \frac{\sqrt{3}}{2} - \sqrt{3}x_G
\end{cases}.
$$
As said, $y_R = y_Y$, indeed:
$$y_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_B = \frac{\sqrt{3}}{2} - \sqrt{3}x_G = y_Y.$$
Moreover, $h = y_R$, then:
$$h = \frac{\sqrt{3}}{2} - \sqrt{3}x_G.$$
Finally, we found that the rectangle is defined by:
$$\begin{cases}
b = 2x_G\\
h = \frac{\sqrt{3}}{2} - \sqrt{3}x_G
\end{cases},$$
while its area is
$$ A =bh = 2x_G\left(\frac{\sqrt{3}}{2} - \sqrt{3}x_G\right).$$
Your rectangle is a square only when $b=h$. That is, when:
$$2x_G = \frac{\sqrt{3}}{2} - \sqrt{3}x_G \Rightarrow x_G = \frac{\sqrt{3}}{2(2+\sqrt{3})}.$$
Then:
$$b = h = \frac{\sqrt{3}}{2+\sqrt{3}}$$
and the area is $$A = \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^2 = \frac{3}{7+4\sqrt{3}}.$$
It is very important to notice that this square is unique possible by construction. Indeed, we started from a rectangles, and we "proved" that this rectangle is a square only if $b=h= \frac{\sqrt{3}}{2+\sqrt{3}}$. So, the maximum area is 
$$A =\frac{3}{7+4\sqrt{3}} = 21 - 12 \sqrt{3} \simeq 0.215390309173472.$$
