Some smoke and mirrors here.
Quoting... "Take the case of a single-phase, 15 amp lighting load operating continuously. To simplify, assume the load is concentrated 100 ft. from the panel"
The maths following looks correct, but makes a huge assumption, namely the load current remains at 15A. In the case of a real-world such as an incandescent lamp, the current consumed by the lamp is related to the voltage across it. So if you upsize your conductors, the I2R losses diminish, the voltage drop across the conductors reduces, so the voltage across the lamp increases, so the lamp consumes more current, so you dont save electricity - you use more of it!!! And (edited to note) since incandescent lamp life changes by (so I've been told...) the twelfth power of the voltage you'll be replacing lamps more often too - more money in the trash can.
Whats that smell around here? Oh yeah, someone let the cattle in...
Though to be fair, if the load was a PC, or a fluoro with electronic "ballast", then these class of device use the same power (KVA) irrespective of the supply voltage, so bigger conductors will save you money for that kind of load.
Better than bigger conductors is higher voltage to reduce the current, along with smaller conductors. I2R losses go up by the square of the current, so the conductor losses for a 10A 120V load (1200W) are four times as high as for a 5A 240V load also 1200W. Always use the highest voltage you can get away with.