Maximize Power Delivered to Circuits
Optimization Problem

The first derivative is used to maximize (optimize) the power delivered to a load in electronic circuits.

Problem

Solution to the Problem

We first express power $P$ in terms of $E$, $r$, and the variable $R$ by substituting i = $\dfrac{E}{(r + R)}$ into P = $R i^2$.
$$\text{P(R)} = \frac{R E^2}{(r + R)^2}$$
We now differentiate $P$ with respect to the variable $R$ $$\frac{dP}{dR} = \frac{E^2 [(r + R)^2 - R 2 (r + R)]}{[(r + R)^4]} = \frac{E^2 [(r + R) - 2 R]}{(r + R)^3} = \frac{E^2 (r - R)}{(r + R)^3}$$ To find out whether $P$ has a local maximum, we need to find the critical points by setting $\dfrac{dP}{dR} = 0$ and solve for $R$.
Since $r$ and $R$ are both positive (resistances), $\dfrac{dP}{dR}$ has only one critical point at $R = r$. Also for $R \lt r$, $\dfrac{dP}{dR}$ is positive and $P$ increases, and for $R > r$, $\dfrac{dP}{dR}$ is negative and $P$ decreases. Hence $P$ has a maximum value at $R = r$. The maximum power is found by setting $R = r$ in $P(R)$
$$\text{P(r)} = \frac{r E^2}{(r + r)^2} = \frac{E^2}{4r}$$
So in order to have maximum power transfer from the electronic circuit to the load $R$, the resistance of $R$ has to be equal to $r$.
As an example, the plot of $P(R)$ for $E = 5$ volts and $r = 100$ Ohms is shown below and it clearly shows that $P$ is maximum when $R = 100$ Ohms = $r$.

References and Links