impedance match with rectifier
You have confused your requirements. The rule "load impedance equals source impedance for maximum load power" applies when the source impedance is fixed. In your case you can reduce the source impedance, so you need to consider other issues. Let's say your power supply has a 10 volts output with a .01 ohm impedance. Then the maximum load power will indeed occur with a load impedance of .01 ohms. However, the output current will be $$i = \frac{V}{2R} = \frac{10}{.02}= 500\text{ amps}$$ and the total power dissipated will be $$P = Vi = 500\text{ amps}\times 10\text{ volts} = 5000\text{ watts}$$ which is not exactly a useful number for a power supply, since it says that the power supply will dissipate 1 watt for every watt in the load, and the power supply will be very inefficient.

Instead, what you need to look at is minimizing the amount of power dissipated in the power supply, and here your concerns about the 40 ohms is perfectly correct. Let's say your rectifier (with an output capacitor to give something like DC) puts out 10 volts, and your load resistance is 10 ohms. Since the output resistance is not fixed, what is the effect of changing it? The answer should be obvious. The lower the source resistance the higher the load voltage and the higher the load power.

So, yeah, get rid of the 40 ohms. It's not helping.