Roy Lewallen wrote:
But what happens if we reduce the source resistance to 50 ohms?
Now the source delivers 66.7 watts, of which 22.2 is dissipated
in the source resistance and 44.4 in the load resistance.

The well known and often misapplied rule about maximizing power transfer
by matching the source and load impedances applies only when you're
stuck with a fixed source resistance and can only modify the load.

What if we assume that we are stuck with the 50 ohm source
impedance above? How do we get greater than 50% power transfer?
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73, Cecil http://www.w5dxp.com