1. Given that $a, b, x, y$ are positive real numbers, satisfying:
$$
a+b+\frac{1}{a}+\frac{9}{b}=8, a x^{2}+b y^{2}=18 \text {. }
$$

Then the range of values for $a x+b y$ is . $\qquad$
\begin{array}{l}\text { I. 1. }(3 \sqrt{2}, 6 \sqrt{2}] . \\ \text { From } 8=a+b+\frac{1}{a}+\frac{9}{b} \\ =\left(a+\frac{1}{a}\right)+\left(b+\frac{9}{b}\right) \\ \geqslant 2+2 \sqrt{9}=8 \\ \Rightarrow a=1, b=3 \\ \Rightarrow \frac{x^{2}}{18}+\frac{y^{2}}{6}=1 . \\ \text { Let } x=3 \sqrt{2} \cos \theta, y=\sqrt{6} \sin \theta\left(0<\theta<\frac{\pi}{2}\right) .\end{array}
\begin{array}{l}\text { Therefore } a x+b y=x+3 y \\ =3 \sqrt{2} \cos \theta+3 \sqrt{6} \sin \theta \\ =6 \sqrt{2} \sin \left(\theta+\frac{\pi}{6}\right) \in(3 \sqrt{2}, 6 \sqrt{2}] .\end{array}