Answer
The maximum value of the power is $\frac{E^2}{4r}$
Work Step by Step
We can find the value of $R$ where $P'(R) = 0$:
$P(R) = \frac{E^2R}{(R+r)^2}$
$P'(R) = \frac{E^2(R+r)^2-2E^2R(R+r)}{(R+r)^4} = 0$
$E^2(R+r)^2-2E^2R(R+r) = 0$
$E^2(R+r)^2=2E^2R(R+r)$
$(R+r)=2R$
$R = r$
$P'(R) = \frac{E^2(R+r)}{{(R+r)^4}}\cdot (r-R)$
When $0 \lt R \lt r$, then $P'(R) \gt 0$
When $R \gt r$, then $P'(R) \lt 0$
Thus, $R=r$ is the value of $R$ where the power $P$ is a maximum.
We can find the maximum power:
$P = \frac{E^2R}{(R+r)^2}$
$P = \frac{E^2r}{(r+r)^2}$
$P = \frac{E^2r}{4r^2}$
$P = \frac{E^2}{4r}$
The maximum value of the power is $\frac{E^2}{4r}$
