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
In the electronic circuit shown below, the voltage E (in Volts) and resistance r (in Ohms) are constant. R is the resistance of a load. In such a circuit, the electric current i is given by
i = E / (r + R)
and the power P delivered to the load R is given by
P = R i^{ 2}
r and R being positive, determine R so that the power P delivered to R is maximum.
Solution to the Problem

We first express power P in terms of E, r and the variable R by substituting i = E / (r + R) into P = R i^{ 2}.
P(R) = R i^{ 2} = R E^{ 2} / (r + R)^{ 2}

We now differentiate P with respect to the variable R
dP / dR = E^{ 2} [ (r + R)^{ 2}  R 2 (r + R) ] / [(r + R)^{ 4} ]
= E^{ 2} [ (r + R)  2 R ] / [(r + R)^{ 3} ]
= 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
dP / dR = 0 and solve for R.

Since r and R are both positive (resistances) dP / dR has only one critical point at R = r. Also for R < r, dP / dR is positive and P increases and for R > r, 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)
P(r) = r E^{ 2} / (r + r)^{ 2} = 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.

Let us examine P(R) again. If R approaches zero, P(R) also approaches zero. If R increases indefinitely, P(R) approaches zero since the horizontal asymptote of the graph of P(R) is the horizontal axis. So that somewhere for a finite value (found to be r) P(R) has a maximum value.
More references on
calculus problems
