Well of course your math is absolutely correct.
And of course, even if you raise the voltage, you increase the current pushed through the resistance, which delivers more current to the battery but also wastes proportionately more watts in the wire. In any event, the less resistance in the wire, the more current delivered to the battery.
Someone mentioned using a 3 wire extension cord to deliver a two wire solution. If we simply short (the same) two of the three conductors at each end, then the resistance in one leg is halved - assuming equivalent gauge in all three wires. Thus the total resistance is 1.5 R instead of 2 R.
All of which shows why we try to use the right wire for the job, but as the OP pointed out, simply getting the panel in the sun over a 50' lossy extension cord will almost certainly be more efficient (deliver more current anyway) than the same panel in the shade over lossless wire.
Also I am looking at higher current fold-able panels which I certainly wouldn't want to hookup over a 16 gauge wire.