Power Matching

To find the maximum power dissipated in the resistor, $P$, we need to find an expression for $P$ in terms of $R_L$, and then differentiate. Note that finding the maximum of $P$ is equivalent to finding the minimum of $\frac{1}{P}$, which is an easier calculation in this case.
$$ I = \frac{V}{R+R_L} $$ $$ P = VI = V^2 \frac{R_L}{(R+R_L)^2} $$ $$ \frac{1}{P} = \frac{1}{V^2} \left(\frac{R^2 + 2RR_L + R_L^2}{R_L}\right) $$ $$ \frac{1}{P} = \frac{1}{V^2} \left(\frac{R^2}{R_L} + 2R + R_L\right) $$ $$ \frac{d}{dR_L} \left(\frac{1}{P}\right) = \frac{1}{V^2} \left(-\frac{R^2}{R_L^2}+1\right)=0 $$ $$ 1 - \frac{R^2}{R_L^2} = 0 $$ $$ R_L = R $$
Therefore to maximise the power across the load, the load resistor should be equal to the resistor $R$. This effect is called 'Power Matching'.