2. Let planar vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}|=3$. Then the maximum value of $a \cdot b$ is $\qquad$
2. $\frac{9}{4}$.

Notice that,
$$
\boldsymbol{a} \cdot \boldsymbol{b}=\frac{1}{4}\left((\boldsymbol{a}+\boldsymbol{b})^{2}-(\boldsymbol{a}-\boldsymbol{b})^{2}\right) \leqslant \frac{9}{4},
$$

the equality holds if and only if $\boldsymbol{a}=\boldsymbol{b}$. Therefore, the maximum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\frac{9}{4}$.