6. $[\mathbf{5}]$ Find the maximum value of $x+y$, given that $x^{2}+y^{2}-3 y-1=0$.
Answer: $\frac{\sqrt{26}+3}{2}$ We can rewrite $x^{2}+y^{2}-3 y-1=0$ as $x^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{4}$. We then see that the set of solutions to $x^{2}-y^{2}-3 y-1=0$ is the circle of radius $\frac{\sqrt{13}}{2}$ and center $\left(0, \frac{3}{2}\right)$. This can be written as $x=\frac{\sqrt{13}}{2} \cos (\theta)$ and $y=\frac{\sqrt{13}}{2} \sin (\theta)+\frac{3}{2}$. Thus, $x+y=\frac{3}{2}+\frac{\sqrt{13}}{2}(\cos (\theta)+\sin (\theta))=\frac{3}{2}+\frac{\sqrt{13}}{2} \sqrt{2} \sin \left(\theta+45^{\circ}\right)$, which is maximized for $\theta=45^{\circ}$ and gives $\frac{\sqrt{26}+3}{2}$. (We could also solve this geometrically by noting that if $x+y$ attains a maximum value of $s$ then the line $x+y=s$ is tangent to the circle.)