14. Given real numbers $x, y$ satisfy $2 x^{2}+4 x y+2 y^{2}+x^{2} y^{2}=9$, let $w=2 \sqrt{2}(x+y)+x y$, then the range of $w$ is $\qquad$
14. When $\sin \theta=-1$, $\theta=-\frac{\pi}{2}$, at this time $w=-6$. When $\sin \theta=\frac{1}{3}$, $\theta=\arcsin \frac{1}{3}$, at this time $w=2(1+\sqrt{2})$. $2 x^{2}+4 x y+2 y^{2}+x^{2} y^{2}=9$ can be transformed into $2(x+y)^{2}+(x y)^{2}=9$. Let $x+y=\frac{3}{\sqrt{2}} \cdot \cos \theta, x y=3 \sin \theta$, then $x, y$ are the two roots of the quadratic equation $t^{2}-\frac{3}{\sqrt{2}} \cos \theta \cdot t+3 \sin \theta=0$. By $\Delta=\frac{9}{2} \cos ^{2} \theta-12 \sin \theta=-\frac{9}{2}\left(\sin \theta \div \frac{4}{3}\right)^{2}+$ $\frac{25}{2} \geqslant 0$, thus $-1 \leqslant \sin \theta \leqslant \frac{1}{3}$, hence $w=3 \cos \theta+6 \sin \theta=3 \sqrt{5} \sin \left(\theta+\arctan \frac{1}{2}\right)$.