# How do you find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle?

base=L/2 height=$L \cdot \frac{\sqrt[2]{3}}{4}$
Let the upper base $y$ of the rectangle be the segment of a line parallel to the base of the equilateral triangle at an unknown distance x from it. In such a way the triangle is divided in two triangles, the equilateral one having height $h = L \frac{\sqrt[2]{3}}{2}$ and a smaller one having height ${h}_{1} = L \frac{\sqrt[2]{3}}{2} - x$, that are similar! so we can write the proportion $\setminus \frac{L}{y} = \setminus \frac{L \frac{\sqrt[2]{3}}{2}}{L \frac{\sqrt[2]{3}}{2} - x}$. By insulating the $y$ we obtain $y = L - \setminus \frac{2}{\sqrt[2]{3}} x$
The rectangle area is $S \left(x , y\right) = x \cdot y$ but $S \left(x\right) = x \cdot \left(L - \setminus \frac{2}{\sqrt[2]{3}} x\right) = L x - \setminus \frac{2}{\sqrt[2]{3}} {x}^{2}$.
By deriving $S \left(x\right)$ we get $S ' \left(x\right) = L - \setminus \frac{4}{\sqrt[2]{3}} x$ whose root is $x = L \setminus \frac{\sqrt[2]{3}}{4}$ and consequently $y = L - \setminus \frac{2}{\sqrt[2]{3}} \setminus \frac{\sqrt[2]{3}}{4} L = \setminus \frac{L}{2}$