Finding maximum value from a statement/equation The statement has values of $x$ and $y$ as positive integers: $$\sqrt{x} - \sqrt{11} = \sqrt{y}$$
I have to find the maximum possible values of $\frac{x}{y}$, this what I have done so far:
$$x = (\sqrt{y} + \sqrt{11})^2$$
$$y = (\sqrt{x} - \sqrt{11})^2$$
therefore: $$\frac{x}{y} = \frac{(\sqrt{y} + \sqrt{11})^2}{(\sqrt{x} - \sqrt{11})^2}$$
...
$$\frac{x}{y} = \frac{y + 11 + 2\sqrt{y}\sqrt{11}}{x + 11 - 2\sqrt{x}\sqrt{11}} = \frac{y+\sqrt{y}}{x-\sqrt{x}} + 1$$
and from here I'm not sure what to do...
(Again better title suggestions are also welcome)

Hint: $\frac{x}{y}=\frac{(\sqrt{y}+\sqrt{11})^2}{y}=\frac{y+2\sqrt{y}\sqrt{11}+11}{y}=1+\frac{2\sqrt{y}\sqrt{11}+11}{y}=f(y)$. Find zeros of $f'(y)$ and determine if they are minima or maxima by the second derivative test.
