jmc

Let $x=CA$. Then $\tan \theta =\tan (\angle BAF-\angle DAE)$, and since $\tan \angle BAF=\frac{2\sqrt{3}}{x-2}$ and $\tan \angle DAE=\frac{\sqrt{3}}{x-1}$, we have $$\tan \theta =\frac{\frac{2\sqrt{3}}{x-2}-\frac{\sqrt{3}}{x-1}}{1+\frac{2\sqrt{3}}{x-2}\cdot \frac{\sqrt{3}}{x-1}}=\frac{x\sqrt{3}}{x^2-3x+8}$$ With calculus, taking the derivative and setting equal to zero will give the maximum value of $\tan \theta$. Otherwise, we can apply AM-GM: $$\begin{array}{rl}\frac{x^2-3x+8}{x}=\left(x+\frac{8}{x}\right)-3& \ge 2\sqrt{x\cdot \frac{8}{x}}-3=4\sqrt{2}-3\\ \frac{x}{x^2-3x+8}& \le \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2-3x+8}=\tan \theta & \le \frac{\sqrt{3}}{4\sqrt{2}-3}\end{array}$$ Thus, the maximum is at $\boxed{\frac{\sqrt{3}}{4\sqrt{2}-3}}$.