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?

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 ${\displaystyle h=L\frac{\sqrt[2]{3}}{2}}$ and
a smaller one having height ${\displaystyle h_1=L\frac{\sqrt[2]{3}}{2}-x}$, that are similar! so we can write the
proportion ${\displaystyle \frac{L}{y}=\frac{L\frac{\sqrt[2]{3}}{2}}{L\frac{\sqrt[2]{3}}{2}-x}}$. By insulating the y we obtain ${\displaystyle y=L-\frac{2}{\sqrt[2]{3}}x}$
The rectangle area is ${\displaystyle S(x,y)=x\cdot y}$ but
${\displaystyle S\left(x\right)=x\cdot (L-\frac{2}{\sqrt[2]{3}}x)=Lx-\frac{2}{\sqrt[2]{3}}{x}^2}$
By deriving S(x) we get ${\displaystyle S\prime \left(x\right)=L-\frac{4}{\sqrt[2]{3}}x}$ whose root is ${\displaystyle x=L\frac{\sqrt[2]{3}}{4}}$ and
consequently ${\displaystyle y=L-\frac{2}{\sqrt[2]{3}}\frac{\sqrt[2]{3}}{4}L=\frac{L}{2}}$