Answer
$\frac{L}{2}$ and $\frac{\sqrt {3}L}{4}$
Work Step by Step
The height $h$ of the equilateral triangle with sides of length $L$ is:
$h^{2}+(\frac{L}{2})^{2}$ = $L^{2}$
$h^{2}$ = $L^{2}-(\frac{L}{2})^{2}$ = $\frac{3L^{2}}{4}$
$h$ = $\frac{\sqrt 3{L}}{2}$
Using similar triangles we have:
$\frac{\frac{\sqrt 3{L}}{2}-y}{x}$ = $\frac{\frac{\sqrt 3{L}}{2}}{\frac{L}{2}}$
$\frac{\frac{\sqrt 3{L}}{2}-y}{x}$ = $\sqrt 3$
$y$ = $\frac{\sqrt 3{L}}{2}-\sqrt {3}x$
The area of the inscribed rectangle is
$A(x)$ = $(2x)y$ = $\sqrt {3}x(L-2x)$ where $0$ $\leq$ $x$ $\leq$ $\frac{L}{2}$
$A'(x)$ = $\sqrt {3}L-4{\sqrt {3}}x$
$A'(x)$ = $0$
$\sqrt {3}L-4{\sqrt {3}}x$ = $0$
$x$ = $\frac{L}{4}$
$A(0)$ = $A\left(\frac{L}{2}\right)$ = $0$
The maximum occurs when
$x$ = $\frac{L}{4}$
$y$ = $\frac{\sqrt {3}L}{2}-\frac{\sqrt {3}L}{4}$ = $\frac{\sqrt {3}L}{4}$
so the dimensions are $\frac{L}{2}$ and $\frac{\sqrt {3}L}{4}$.
