OK, after fiddling around with various ways of doing this, I think I've come up with a way of explaining what I said above.
I'll use, v = airspeed, w = windspeed, t = time, d = distance, p = power.
Starting with the obvious t = d/v
Now power is related to fuel flow, so the time we have availble is proportional to fuel/power. So t = 1/p for one 'unit' of fuel
Putting them together gives d/v = 1/p, which becomes d = v/p
So for maximum d, we have to maximise v/p, or minimise p/v.
i.e. the usual tangent on the power/velocity graph.
Similarly, if we have a wind then t = d/(v+w)
So if we do the same as above we get, d/(v+w) = 1/p
then d = (v+w)/p
So we have to minimise p/(v+w), or in other words, you have to draw a tangent from the graph to -w on the speed axis.
I hope that's right, or I've completely misunderstood it as well
