7. For any $x, y \in [0,1]$, the function
$$
f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}
$$

has a maximum value of $\qquad$ .
7. 1.

Since $x, y \in [0,1]$, then $x \leqslant \sqrt{x}, y \leqslant \sqrt{y}$.
Let $x=\sin ^{2} \alpha, y=\sin ^{2} \beta\left(\alpha, \beta \in\left[0, \frac{\pi}{2}\right]\right)$.
Thus, $f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}$
$$
\begin{array}{l}
\leqslant \sqrt{x(1-y)}+\sqrt{y(1-x)} \\
=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta \\
=\sin (\alpha+\beta) \leqslant 1,
\end{array}
$$

The equality holds if and only if $\alpha+\beta=\frac{\pi}{2}$, and
$$
x=\sqrt{x}, y=\sqrt{y} \text {. }
$$

At this point, $\{x, y\}=\{0,1\}$.