I wanted to know what the load resistance should be to achieve the highest power consumption at said load given a source voltage and its ESR (Equivalent series resistance). Simple, right? Well, after some number-herding I found that the power at the load is equal to (V^2*ESR)/(ESR+Load)^2.

To find where it's maximum lies, we want to find points in which it's slope is zero, in other words, find the point where the differential equals 0. So, I cheated and used an online calculator to find the differential, which so happened to be THIS sucker:

-(load-ESR)*V^2/(load^3+3*ESR*load^2+3*ESR^2*load+ESR^3)

I then set it equal to 0, and solved for the load. (That is, what does the load have to be to set this entire thing to 0, which, again, is where the maximum point lies in my power equation). Turns out it's load = ESR

That's it. load equals ES-effing-R.

Now that I see that, I realize it wasn't that freaking hard to come by, it just didn't occur to me earlier intuitively. I was graphing crap, flipping papers back and forth, scribbling and circling equations and garbage for like half an hour. It was still really fun, and it's awesome to know that, even though I complicated the HELL out of this, the solution still came out ok, so at least I did the complicated crap correctly...

To solve a bit more intuitively, when load = ESR, 1/2 of the total power consumption is in the load (not very efficient, but regardless). Also, when load = ESR, the total power consumption is 1/2 what it could have been if the load resistance was zero. So, the load power is equal to the maximum power output divided by 4, or Maximum-Load-Power = V^2/(ESR*4). What does the load have to be to get that? Well, the voltage at the load is Vs*load/(load+ESR) (like a voltage divider), so:

(Vs*load/(load+ESR))^2/load = Vs^2/(ESR*4) solve for load, and load = ESR

I just thought this was very interesting. Some other ideas might be "where is the most efficient operating point?, that is, power-Load/power-Loss" or something similar.

Are there any simpler solutions to this problem that anyone can find? the simpler, the more likely we'll "see" or "feel" its solution, which leads to it's mastery.