Rotation angle of regular polygon that has largest taxicab maginitude between all vertices Firstly just to apologise, I posted this on mathoverflow before realising it was focused on research level mathematics. 
If I have a regular polygon that is centred at the origin. Then take the difference between all vertices.
Let $U$ = the maximum $x$ difference 
Let $V$ = the maximum $y$ difference
I want to figure out the angle of rotation that creates the maximum 
$$W = U + V \, .$$
So for example imagine a line (I know it's not a polygon); if the angle of rotation is $0$ then its vertices are at $( 1, 0 )$ and $( -1, 0 )$
$$
U = 1 - (-1) = 2\\
V = 0 - 0 = 0
$$
So $W = 2 + 0 = 2$.
The optimum is 45 degrees of rotation in which case 
$$
U = 0.707 - -0.707 = 1.414 \\
V = 0.707 - -0.707 = 1.414
$$
So $W = 2.828$.
If my calculations are correct, an equilateral triangle's optimum is $45$ degrees as well. However its obvious to see the optimum for a square is 0 since the vertices align perfectly with $x,y$ axis (i.e., $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$).
Hopefully I've explained it well enough. I can find the solution for single instance of the problem using graphs and derivatives BUT it takes a long time. I'm looking for a general purpose algorithm as this is going to be used in some software I'm writing.
Thanks in advance.
Iain
PS I forgot to mention that there'll be multiple solutions due to symmetry. So the line's optimum is $45, 135, 225, 315$. I believe the triangles is $15 + 30n$, i.e. $15, 45, 75, \dots$
PPS The vertices are generated using the following formula:
Let $n$ = number of sides
Let $c = 360 / n$
For all  $i:    0  \le  i  \lt  n$ 
$V_i  = ( \cos(i  \cdot  c ), \sin( i \cdot  c ) )$
E.g. a regular triangles vertices are: 
$V_0  =(\cos(0^\circ), \sin(0^\circ) ) = (1, 0)$
 $V_1  =(\cos(120^\circ), \sin(120^\circ) ) = (-0.5, 0.866)$
 $V_2  =(\cos(240^\circ), \sin(240^\circ) ) = (-0.5, 0.866)$
 A: You need to define what the angle of rotation is measured from.  I would have taken zero rotation for the square to be with the sides parallel to the axes.  Having a corner on the axes is clearly optimal for $4n$ sided polygons.  For $4n+2$ sided polygons it will not be optimal because you gain more from rotating the flat sides than you lose by rotating the points.  
To be exact, let $0$ represent having points on the $x$ axis.  We start with points $(1,0), (\cos \frac {2n}{4n+2} \pi ,\sin \frac {2n}{4n+2}\pi),(-1,0), (-\cos \frac {2n}{4n+2} \pi ,-\sin \frac {2n}{4n+2}\pi)$ and if we rotate counterclockwise by $\theta$ we get $U=2 \cos \theta,V=2\sin (\frac {2n}{4n+2}\pi+\theta)$  We add these up, differentiate, set to zero, etc.
$$U+V=2 \cos \theta +2\sin \left(\frac {2n}{4n+2}\pi+\theta\right)\\\frac d{d\theta}(U+V)=-2 \sin \theta+2\cos \left(\frac {2n}{4n+2}\pi+\theta\right)\\0=-2 \sin \theta +2\cos\left(\frac {2n}{4n+2}\pi \right) \cos \theta-2\sin\left( \frac {2n}{4n+2}\pi\right)\sin \theta\\\tan \theta=\frac{\cos\left(\frac {2n}{4n+2}\pi \right)} {\sin\left( \frac {2n}{4n+2}\pi\right)+2}$$
