11. Given acute angles $\angle A, \angle B$ satisfy $\tan (A+B)=2 \tan A$.
Then the maximum value of $\tan B$ is $\qquad$
11. $\frac{\sqrt{2}}{4}$.

Notice,
$$
\begin{array}{l}
\tan B=\tan [(A+B)-A] \\
=\frac{\tan (A+B)-\tan A}{1+\tan (A+B) \cdot \tan A} \\
=\frac{2 \tan A-\tan A}{1+2 \tan ^{2} A}=\frac{\tan A}{1+2 \tan ^{2} A} \\
\leqslant \frac{\tan A}{2 \sqrt{2} \tan A}=\frac{\sqrt{2}}{4} .
\end{array}
$$

Equality holds if and only if $\tan A=\frac{\sqrt{2}}{2}$. Therefore, the maximum value of $\tan B$ is $\frac{\sqrt{2}}{4}$.