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?

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?
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Makenna Lin
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 $\frac{L}{y}=\frac{L\frac{\sqrt[2]{3}}{2}}{L\frac{\sqrt[2]{3}}{2}-x}$. By insulating the y we obtain $y=L-\frac{2}{\sqrt[2]{3}}x$
The rectangle area is $S\left(x,y\right)=x\cdot y$ but
$S\left(x\right)=x\cdot \left(L-\frac{2}{\sqrt[2]{3}}x\right)=Lx-\frac{2}{\sqrt[2]{3}}{x}^{2}$
By deriving S(x) we get $S\prime \left(x\right)=L-\frac{4}{\sqrt[2]{3}}x$ whose root is $x=L\frac{\sqrt[2]{3}}{4}$ and
consequently $y=L-\frac{2}{\sqrt[2]{3}}\frac{\sqrt[2]{3}}{4}L=\frac{L}{2}$