Question:
If a rectangle is inscribed in an equilateral triangle of side length $2 \sqrt{2}$ as shown in the figure, then the square of the largest area of such a rectangle is__________.
Solution:
In $\triangle \mathrm{DBF}$
$\tan 60^{\circ}=\frac{2 b}{2 \sqrt{2}-\ell} \Rightarrow b=\frac{\sqrt{3}(2 \sqrt{2}-\ell)}{2}$
$\mathrm{A}=$ Area of rectangle $=\ell \times \mathrm{b}$
$A=\ell \times \frac{\sqrt{3}}{2}(2 \sqrt{2}-\ell)$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \ell}=\frac{\sqrt{3}}{2}(2 \sqrt{2}-\ell)-\frac{\ell \cdot \sqrt{3}}{2}=0$
$\ell=\sqrt{2}$
$A=\ell \times b=\sqrt{2} \times \frac{\sqrt{3}}{2}(\sqrt{2})=\sqrt{3}$
$\Rightarrow \mathrm{A}^{2}=3$