Yes but so is the load. Yes it is how it is done.
Hmm. The output impedance of a good quality audio amplifier seems very low because it has a negative feedback loop which makes it makes it appear so . Far less than the nominal speaker impedance - typically 8 ohms. I used to design audio for Cambridge Audio back in the 70's, and I think we quoted 'damping factors' of > 100 - ie an output impedance of < 8/100 ohms over a large frequency range.
The actual open loop output impedance and hence overall power transfer efficiency has to take into account the psu impedance as well as the output devices - and gets nowhere near the maximum power transfer point, or things would burn up! ( and can if you short the speaker out on an amp with inadequate protection circuitry.)
Typical scenario.......
Say the psu has a ,< 1/2 ohm Zout, big BJT's or mosfets <<1 ohm. hard on ... 60v rail....max power transfer point would occur into that load impedance ( >100A) , and the amp wouldn't be happy!
Back to normal operation.........
At any power the resistance of the power devices is continuously adjusted so as to deliver the required current into the load. At low powers the efficiency is very poor. Best efficiency ( for a class B amp ) occurs at flat out, ie p-p = rail-to-rail ...... 70 odd percent ? ( I'd have to redo my sums to confirm) for a sinewave. At some point in that power output level there will be maximum power transfer (50% effcy) into the eg 8 ohm load - but it will be transient.
but then efficiency wasn't what we were optimising.
In class A amps ( eg Sugdens) the amp is hottest when delivering least power..... We didn't go that way.
regards
Allan
ps - sorry, ilovetoflyfpv - back to your question. The circuitry you propose will be fine.
Hi allanhurst and Grumpy_Mike
Should I be concerned about the resistor selection for the voltage divider achieving 10kΩ total resistance? I noticed in the Read Analog Voltage Tutorial a 10kΩ potentiometer is used which I imagine was selected because it would draw little current (0.5mA). Is this an upper limit?
Alternatively, should my focus be on the capacitor resistor combination achieving 10kΩ impedance? I noticed in the Analog Input Tutorial a 10kΩ potentiometer is used again but this time reference is made to the high impedance of the analog pins. Is there a current limit here or something else I need to pay attention to?
As always appreciate your help.
Cheers
Jase 
A potentiometer allows you to deal with a larger range of signal levels. In your original post you said the the level was 0.5 - 2.0 volts. average? peak-peak? rms?
A sine wave with 2v rms will have a peak-peak value of 2 x 2 x root(2) or 5.65 volts - too much for the 5v range of the arduino - but the 10k/ 50k effective attenuator of the given circuit will reduce this to 50/(50+10) times ie giving about 4.7 volts p-p , which is OK. Don't bother with the pot.
Just build it, try it, and get coding!
regards
Allan
imagine was selected because it would draw little current (0.5mA). Is this an upper limit?
There is no upper limit except that determined by the pot. You can have what ever current you want. However as you are learning things are a balance, so a 10K pot is a balance between not using too much current for no reason, and being able to charge thi thump capacitor resasionably fast.
Alternatively, should my focus be on the capacitor resistor combination achieving 10kΩ impedance?
In practice you should use two 20K resistors and a capacitor that doesn't suffer much with that load. You are trying to maximise voltage transfer, so make the capacitor as big as convenient. You are trying to overthink this problem, things are not as critical as you think they are. It is fine knowing the theory but you have to choose a value sooner or later. I used to tell my students that electronics is full of calculations and the choosing a value of component in the same order of magnitude.
To paraphrase Bob Pease - a great engineer - on PSPICE - a simulation program, which he despised -
'my simulation tool is a breadboard and a soldering iron'
sometime or other you have to stop calculating and guessing and build it.
regards
Allan
Hi allanhurst, Grumpy_Mike and MarkT
Thanks so much for your help. Once the dude in the big red suit heads back to the North Pole I'll hopefully get a chance to apply what I've learnt.
Cheers
Jase