Calculus in a simple resistance problem

The problem goes as follows: a resistor, resistance R, is connected in series with a battery, emf ε and internal resistance r; find the maximum power dissipation in the resistor R, and the resistance that results in this output.

Although it is quite clear how this can be solved with algebra, some thought about the theory behind Ohm's Law and Kirchoff's Second Law is required to set the answer in some context: since the power dissipation in the resistor is equal to I²R, one would think that raising the current is more influential on the output than raising the resistance. However, since the current drawn from the battery is intrinsically linked to the total resistance of the circuit, there is a delicate balance to be struck.
The nature of this balance is difficult to ascertain without calculation though, so without further ado:

The result is that the power dissipated across the resistor is a maximum (assumed from the context of the function whose stationary point is found, since only one positive-R stationary point exists) when R = r. This nicely helps to explain why a short-circuited battery is likely to heat up, set fire or explode - the short-circuit has a very low resistance, so overall a massive current is drawn which causes the wire to heat up. The wire will heat up the most when the resistance of the wire matches the internal resistance of the battery. The overall implication is that, if one was designing an electric resistance heater to emit the maximum heat output for any given emf, the length of wire coiled up should, via the resistivity equation, reflect the internal resistance of the power supply.

However, a separate issue is maximising the efficiency of the circuit in terms of energy - for environmental and financial reasons it is always of high priority to do this in power-hungry domestic appliances, and the heater is no exception. Since the efficiency of the circuit is the ratio of power dissipation in the battery to the total power dissipation of the circuit, maximum efficiency would be achieved when the resistance of R is infinitely high. The issue with this is, just like trying to connect a 12V power pack across a piece of plastic, you cannot expect any power to be generated at all - this is because the high resistance results in a tiny current being drawn from the battery and in any case, the power output would be highly suboptimal according to the result I derived above. The moral of the story for the heater is that the most important factor here is to cut the internal resistance of the power supply as much as possible.

If the heater is being connected to the mains power, this opens an entirely new can of worms, namely the definition of the National Grid's internal resistance. This is a combination of the cumulative resistance of electrical cabling across the country, inefficiencies in combustion of fuels in power stations, friction in turbine halls, and delayed responses in adjusting power distribution across the country following surges and drops in energy demand, amongst other more subtle factors. Nevertheless more simple domestic factors to consider are reducing usage of extension leads where possible, and keeping the heater cable well insulated so that its resistance doesn't itself rise with time!