GS. 2 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 $d$.
$\begin{array}{l}x+y=(\sqrt{x}+\sqrt{y})^{2}-2 \sqrt{x y} \\ \Rightarrow(\sqrt{x}+\sqrt{y})^{2}=18+2 \sqrt{x y} \leq 18+2\left(\frac{x+y}{2}\right)=36 \quad \text { (G.M. } \leq \text { A.M.) } \\ \sqrt{x}+\sqrt{y} \leq 6=d\end{array}$