the range of the vector dot product $\overrightarrow{P M} \cdot \overrightarrow{P N}$ 
$ABCDEF$ is a equilateral regular hexagon with $AB=2 \sqrt2$.  Let $P$ be an arbitrarily point of the hexagon and  $MN$ be a movable chord of the circumcircle of  the  hexagon with $MN=4$.  What would the range of  the vector dot product $\overrightarrow{P M} \cdot \overrightarrow{P N}$ be?

My attemption: 
Set up an coordinate system as following: 
Let the coordinate of $P$ be  $(x,-\sqrt6)$  with  $-\sqrt2 \leq x \leq \sqrt2$, and set the coordinate values of the points $M, N$ to be  $(2\sqrt2 \cos \alpha,2\sqrt2 \sin \alpha), (-2\sqrt2 \sin\alpha,2\sqrt2 \cos \alpha)$ respectively, where $0 \leq \alpha \leq 2\pi$. Thereby, we obtain
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=x^2+4x\sin(\alpha-\frac{\pi}{4})+4\sqrt6\sin(\alpha+\frac{\pi}{4})+6.$$
Set $t=\alpha-\frac{\pi}{4}$, then 
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=x^2+4x\sin t+4\sqrt6\cos t+6,$$
for $0 \leq t \leq 2\pi$ and $-\sqrt2 \leq x \leq \sqrt2$.
On one hand,
\begin{align*}
x^2+4x\sin t+4\sqrt6\cos t+6 \leq& 2+4(\sqrt2|\sin t|+\sqrt6 |\cos t|)+6\\
&=8+4(\sqrt2|\sin t|+\sqrt6 |\cos t|)\\
&\leq 8+8\sqrt2.
\end{align*}
Obviously, the maximum value is $8+8\sqrt2$.
How do I proceed ahead now? Any help would be appreciated.

Continue with
$$\overrightarrow{P M} \cdot \overrightarrow{P N}=
f(x, t)=x^2+4x\sin t+4\sqrt6\cos t+6,$$
Note that $f_x’=f_t’=0$ yields $(x,t)=(0,0),(0,\pi),(0,2\pi)$, which leads to extreme values $6\pm 4\sqrt6$. 
Moreover, check the boundary values at $x=\pm\sqrt2$,
$$f(\pm\sqrt2,t)= 2\pm4\sqrt2 \sin t+4\sqrt6 \cos t+6
=8\pm8\sqrt2\sin(t+\theta)$$
Together, the range is 
$$6-4\sqrt6 \le \overrightarrow{P M} \cdot \overrightarrow{P N}\le 8+8\sqrt2$$
