Calculate the value of expression $Q = \frac{x + 1}{y}$ when $xy > 1$ and expression $P = x + 2y + \frac{5x + 5y}{xy - 1}$ reaches its maximum value. 
Consider two positives $x$ and $y$ where $xy > 1$. The maximum value of the expression $P = x + 2y + \dfrac{5x + 5y}{xy - 1}$ is achieved when $x = x_0$ and $y = y_0$. Calculate the value of expression $Q = \dfrac{x_0 + 1}{y_0}$.

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Isn't being demotivated one of the worst feelings in the world?
Let $x = \cot\alpha \left(n\pi < \alpha < n\pi + \dfrac{\pi}{2}, n \in \mathbb Z\right)$ and $y = \cot\beta \left(n\pi < \beta < n\pi + \dfrac{\pi}{2}, n \in \mathbb Z\right)$. It could be implied that $P = \cot\alpha + 2\cot\beta + \dfrac{5}{\cot(\alpha + \beta)}$.
Hmmm~ never-mind, this leads to nowhere.
Perhaps creating new variables, specifically $a = x + 1 \ (a > 2)$ and $b = \dfrac{1}{y} \ (b > 0)$ might actually be a clue. Then, we have that $a - 1 > 2b$ and $$\begin{aligned} P = a + 5b - 1 + \dfrac{2}{b} + \dfrac{5(b^2 + 1)}{a - b - 1} &= 6b + \dfrac{2}{b} + (a - b - 1) + \dfrac{5(b^2 + 1)}{a - b - 1}\\ &\ge 2\left[3b + \sqrt{5(b^2 + 1)} + \dfrac{1}{b}\right] \end{aligned}$$
Function $f(b) = 3b + \sqrt{5(b^2 + 1)} + \dfrac{1}{b}$ contains its first derivative as $f'(b) = 3 + \dfrac{b\sqrt{5}}{\sqrt{b^2 + 1}} - \dfrac{1}{b^2}$ and the equation $f'(b) = 0$ has roots $b = \dfrac{1}{2}$ and $b = -\sqrt{\dfrac{\sqrt 5 - 1}{2}}$
We can then draw the table of variations for function $f(b)$.

As $f(b) \ge 6, \forall b > 0$, it can be inferred that the minimum value of $P$ is $2 \times 6 = 12$, achieved when $$\left\{ \begin{aligned} a > 0, b > 0&, a - 1 > 2b\\ (a - b - 1)^2 &= 5(b^2 + 1)\\ b &= \dfrac{1}{2} \end{aligned} \right. \iff \left\{ \begin{aligned} a &= 4\\ b &= \dfrac{1}{2} \end{aligned} \right. \implies Q = ab = 4 \times \dfrac{1}{2} = 2$$
Wait... minimum? That's not what the question asks for, although that's one of the given options.
Anyhow, I'm still figuring out how to tackle this problem. Should I give in to the asymmetry or otherwise? As always, thanks for reading, (and even more so if you could help), have a wonderful tomorrow~
By the way, the options are $\sqrt 2, \sqrt 3, 2$ and $1$, so I could have been fooled there.
 A: Just note that since the points $xy=1$ are excluded, the maximizers/minimizers will be a stationary points in the open set defined by the condition $x y>1$. Once you determine the stationary points (some are complex numbers, rule them out), you'll see that the only stationary point with $x,y>0$ and $xy>1$ is $(x,y)=(3,2)$. Computing the Hessian matrix, it is in fact true that it is a minimizer.
It was clear right from the start that there could not be a maximizer... when we approach the line $xy = 1$ function values tend to $+\infty$.
A: With $P = x+2y+5\frac{x+y}{xy-1}$ and $Q=\frac{x+1}{y}$ and forming the lagrangian
$$
L(x,y,\lambda,s) = P+\lambda(x y-1-s^2)
$$
the stationary points are the solutions for
$$
0 = \nabla L = \cases{
 \lambda  y-\frac{y (5 x+5 y)}{(x y-1)^2}+\frac{5}{x y-1}+1 \\
 \lambda  x-\frac{x (5 x+5 y)}{(x y-1)^2}+\frac{5}{x y-1}+2 \\
 x y-1-s^2 \\
 -2 \lambda  s }
$$
The solutions give us
$$
\left(
\begin{array}{cccccc}
P & Q & x & y & \lambda & s^2\\
 1 & -12 & -3 & -2 & 0 & 5 \\
 2 & 12 & 3 & 2 & 0 & 5 \\
\end{array}
\right)
$$
