As I understand it, should maximum power transfer be when $$ R_L=R_{th} $$ then $$ P_L= P_{L(max)} $$
So that this should do the trick:
$$ P_{L}=\frac{V_{th} ^2}{R_{th}} $$
But it seems that I am doing something wrong.
Electronic – How to find maximum power transfer using Thevenin equivalent
basicthevenin
Related Solutions
Barry gives the correct answer. What follows is the mathematical justification.
The power delivered to the load resistor, for a given Thevenin equivalent circuit is:
\$P_L = V_L \cdot I_L = V_{th}\dfrac{R_L}{R_{th}+ R_L} \cdot \dfrac{V_{th}}{R_{th}+ R_L} = V^2_{th} \dfrac{R_L}{(R_{th}+ R_L)^2}\$
If we fix \$R_{th}\$ and ask which value for \$R_L\$ gives maximum \$P_L\$, the value is given by \$R_L = R_{th}\$ and the resulting power delivered to the load is:
\$P_{L,max} = \dfrac{V^2_{th}}{4R_{th}} \$
However, if we fix \$R_L \$, and ask which value for \$R_{th}\$ gives maximum \$P_L\$, the answer is, by inspection, \$R_{th} = 0\$.
Thus, the answer is \$R = 500 \Omega \$ so that \$R_{th} = 0\$
Redraw the circuit as follows to see why \$V_X\$ is across the 60 ohm resistor, the controlled current source and the output terminals:
It is often helpful to remember that two points connected by an ideal wire are the same circuit node which can be seen by 'shrinking' the wire to zero length as I've done above.
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Best Answer
It's a simple error in your reasoning. When the load resistance is equal to the Thevenin resistance, the voltage across the load is precisely 1/2 the Thevenin voltage.
Thus,
\$P_{L,max} = \dfrac{V^2_{th}}{4R_{th}} \$
Your calculation would be correct if the entire Thevenin voltage appeared across the load but that isn't the case.