Given $x \geqslant 0, y \geqslant 0$, and $x^{2}+y^{2}=1$, then the maximum value of $x(x+y)$ is

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Given $x \geqslant 0, y \geqslant 0$, and $x^{2}+y^{2}=1$, then the maximum value of $x(x+y)$ is
9. $\frac{\sqrt{2}+1}{2}$.

Let $x=\cos \alpha, y=\sin \alpha, \alpha \in\left[0, \frac{\pi}{2}\right]$, then
$$
\begin{aligned}
x(x+y) & =\cos \alpha(\cos \alpha+\sin \alpha) \\
& =\sqrt{2} \cos \alpha \sin \left(\alpha+\frac{\pi}{4}\right) . \\
& =\frac{\sqrt{2}}{2}\left[\sin \left(2 \alpha+\frac{\pi}{4}\right)+\sin \frac{\pi}{4}\right] \\
& \leqslant \frac{\sqrt{2}}{2}\left(1+\frac{\sqrt{2}}{2}\right)-\frac{\sqrt{2}+1}{2} .
\end{aligned}
$$

When $\alpha=\frac{\pi}{8}$, the equality holds. Therefore, the maximum value of $x(x+y)$ is $\frac{\sqrt{2}+1}{2}$.