G3.4 Let $x \geq 0$ and $y \geq 0$. Given that $x+y=18$. If the maximum value of $\sqrt{x}+\sqrt{y}$ is $d$, find the value of $d$. (Reference: 1999 FGS.2)
$$
\begin{array}{l}
x+y=(\sqrt{x}+\sqrt{y})^{2}-2 \sqrt{x y} \\
\Rightarrow(\sqrt{x}+\sqrt{y})^{2}=18+2 \sqrt{x y} \leq 18+2\left(\frac{x+y}{2}\right)=36 \quad(\mathrm{GM} \leq \mathrm{AM})
\end{array}
$$
$\sqrt{x}+\sqrt{y} \leq 6=d$ (It is easy to get the answer by letting $x=y$ in $x+y=18$ )
Remark The original question is Given that $x+y=18$. If the maximum value of $\sqrt{x}+\sqrt{y} \cdots$ $\sqrt{x}+\sqrt{y}$ is undefined for $x<0$ or $y<0$.