2. Given $x, y \in \mathbf{R}, 2 x^{2}+3 y^{2} \leqslant 12$, then the maximum value of $|x+2 y|$ is $\qquad$
Solve $(x+2 y)^{2} \leqslant\left(\frac{1}{2}+\frac{4}{3}\right)\left(2 x^{2}+3 y^{2}\right) \leqslant 22$, when the equality holds, $4 x^{2}=\frac{9}{4} y^{2} \Rightarrow x^{2}=\frac{18}{11}$, $y^{2}=\frac{32}{11}$. Therefore, the maximum value of $|x+2 y|$ is $\sqrt{22}$.