What is the cycle for when the maximum amount of the Moon's apparent area is sunlit? The Moon is defined as full when its ecliptic longitude is 180$^{\circ}$ distant from the Sun's.

But what is the cycle for when the maximum proportion of the moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised?

Notes

1) After first guessing (on this expert Q&amp;A site) that the answer is "one lunar month" and asking me whether there was "a reason" why the said cycle might be different from it, user UhOh has requested clarification of the term "moon's apparent area", pointing out that the moon has a rough surface. The difference of the apparent perimeter of the Moon's disc from a circle does not have a bearing on the answer to this question that is of anything like the magnitude of the contribution made by the 5$^{\circ}$ tilt of the plane of the Moon's orbit around the Earth relative to the plane of the Earth's orbit around the Sun. Please ignore mountains and craters. Please assume either that the Moon is a sphere, or if you wish then take it to be a spheroid with an eccentricity of about 1/900. The question is about the relative movements of the three celestial bodies concerned. (If further clarification is required of any other term used, please let me know.)

2) Wouldn't it be great if there were a website where you could input your latitude and longitude and desired start date and end date and get a graph of the proportion of the moon's apparent disc that is sunlit from your location? But I don't think this exists (yet!)
I asked the same question on the Usenet group sci.astro and received the following helpful reply from Barry Schwarz that goes some way towards an answer, reposted here with permission. But the question remains hard. I have not seen this cycle addressed in any astronomical textbook, although it may be in one that I haven't looked at. 


  Since the Sun's diameter exceeds the Moon's, the Sun always
  illuminates slightly more than 50% of the Moon's surface.  (To
  visualize, draw the external common tangents from a larger circle to a
  smaller one.)  Using 98M miles as the distance, 432K miles as the
  Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost
  7 miles into the opposite hemisphere.
  
  Any observer a finite distance from the Moon can only ever see less
  than 50% of the Moon's surface. The further away the observer is, the
  closer the observable percentage is to 50.  (To visualize, draw the
  two tangents from an external point to a circle.)  However, when the
  Moon is in the plane of the ecliptic, an observer at the north pole
  can see approximately 12 miles over the Moon's north pole into the
  opposite hemisphere.  So he can see some 5 miles beyond the
  terminator.  Similar geometry causes his view of the Moon's south pole
  to fall short of the terminator. 
  
  However, for a person standing in line with the ecliptic, his view in
  any direction falls 4 miles short of the terminator and that is the
  maximum you seek.
  
  When the Moon is within 4K miles of the ecliptic, an observer on Earth
  at the same offset from the ecliptic will have the view described
  above.
  
  When the Moon is more than 4K miles away from the ecliptic, all
  Earth-bound observers see some non-illuminated portion and therefore
  do not see the maximum you want.
  
  Not only is the Moon's orbit tilted with respect to the ecliptic, but
  the direction of the tilt wobbles like a child's top.  I have no idea
  how to factor that into an equation that would compute the  time
  between one visible maximum and another.