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.
 A: 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 
$$6-4\sqrt6 \le \overrightarrow{P M} \cdot \overrightarrow{P N}\le 8+8\sqrt2$$
A: The definition of $P$ is ambiguous, it is not clear whether $P$ is allowed to be an interior point of hexagon or not. This doesn't matter for the maximum but the minimum depends on this.
Let $\mathcal{P}$ be the solid hexagon $ABCDEF$. We will only consider the case $P$ is lying on the sides of the hexagon (i.e $P \in \partial \mathcal{P}$)
Let $Q$ be the midpoint of $MN$. Recall the vector identity $4\vec{u}\cdot\vec{v} = |\vec{u}+\vec{v}|^2 - |\vec{u}-\vec{v}|^2$, we find
$$\overrightarrow{PM}\cdot\overrightarrow{PN} = \left|\frac{\vec{M}+\vec{N}}{2} - \vec{P}\right|^2  - \left|\frac{\vec{M}-\vec{N}}{2}\right|^2 = |PQ|^2 - 4\tag{*1}$$
The problem at hand is equivalent to finding the largest/smallest distance between $P$ and $Q$.
Since $MN$ is a chord of fixed length on a circle, the locus of $Q$ is another circle, let's call it $\mathcal{Q}$, with same center $O$ and radius $\sqrt{(2\sqrt{2})^2 - (4/2)^2} = 2$.
For $\rho < 2$, the locus of points at a distance $\rho$ from $\mathcal{Q}$
forms an annulus with inner radius $2-\rho$ and outer radius $2+\rho$. When $\rho = \sqrt{6} - 2$, the outer rim of this annulus touches the sides of hexagons at their midpoints. This implies 
$$\min_{P\in \partial\mathcal{P},Q \in \mathcal{Q}} |PQ| = \sqrt{6} - 2\tag{*2a}$$
Notice for any point $X$ outside $\mathcal{Q}$, we have
$$\max_{Q\in\mathcal{Q}}|XQ| = |\vec{X}| + 2$$
Since $\partial\mathcal{P}$ lies completely outside $\mathcal{Q}$, we have
$$\max_{P\in \partial\mathcal{P},Q \in \mathcal{Q}} |PQ| 
= \max_{P\in\partial\mathcal{P}}\max_{Q\in\mathcal{Q}}|PQ|
= \max_{P\in\partial\mathcal{P}}|\vec{P}| + 2
= 2\sqrt{2} + 2\tag{*2b}$$
Combine $(*2a), (*2b)$ and apply $(*1)$, we find 
$$6 - 4\sqrt{6} = (\sqrt{6}-2)^2 - 4 \le \overrightarrow{PM}\cdot\overrightarrow{PN} \le (2\sqrt{2}+2)^2 - 4 = 8+8\sqrt{2}$$
