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Just look above figure. It is a practical power source which will have an internal resistance greater than 0.
Now if u want to take maximum voltage at O/P then by using voltage divider law, u will get RL=\infty
In this case, power=0. ({v}^{2 }/infinity )=0
Now if u want maximum current O/P then u can decrease ur RL. right?
So if u decrease it to 0, then u will ger max current as V/Rth. At this time maximum power is delivered across the Rth.
But still , power O/P(power transfer) is 0. ({I}^{ 2} * 0)=0
U could not use {V}^{ 2} /R equation because it is 0/0 form. (V out=0; RL=0)
So the power O/P is zero for RL=0 & RL=infinity.
Now, the power O/P is maximum at a particular value of RL . But it depends on the internal resistance Rth.
For finding the value of RL for maximum ,
u can simply differentiate the Power Output with RL.
Then at peak power value, the rate of change of power with respect to RL will be zero.
So to find it, just substitute
dP/dRL=0;
Now u will get Rth =RL
This means the maximum value of power output will be when Tth = RL.