10. Let $a, b$ be positive constants, $\theta \in\left(0, \frac{\pi}{2}\right)$, find the maximum value of $y=a \sqrt{\sin \theta}+b \sqrt{\cos \theta}$.
10. Let $\sqrt{u}=a \cdot \sqrt{\sin \theta}, \sqrt{v}=b \cdot \sqrt{\cos \theta}$, then the condition transforms to: $\frac{u^{2}}{a^{4}}+\frac{v^{2}}{b^{4}}=1, u$ 、 $v \geqslant 0$, using the Cauchy inequality,
$$\begin{aligned}
y & =\sqrt{u}+\sqrt{v} \leqslant \sqrt{\left(\frac{u}{a^{\frac{4}{3}}}+\frac{v}{b^{\frac{4}{3}}}\right)\left(a^{\frac{4}{3}}+b^{\frac{4}{3}}\right)} \\
& \leqslant \sqrt{\sqrt{\left(\frac{u^{2}}{a^{4}}+\frac{v^{2}}{b^{4}}\right)\left(a^{\frac{4}{3}}+b^{\frac{4}{3}}\right)} \cdot\left(a^{\frac{4}{3}}+b^{\frac{4}{3}}\right)} \\
& =\left(a^{\frac{4}{3}}+b^{\frac{4}{3}}\right)^{\frac{1}{4}}
\end{aligned}$$

It is easy to find the condition for equality to hold.