A constant voltage source V with internal resistance r is connected to a load resistor R. The dissipated power by the resistor R is P=RV^2/(R+r)^2. Show that the maximum power dissipated by the resistor R is achieved when R = r. The maximum of P with respect to R is achieved when dP/dR = 0.
P = RV^2/(R+r)^2
The Attempt at a Solution
Well, I think that I need to find the absolute maximum. That means I need to find the first derivative, find the critical points, evaluate the equation at the critical points and find the absolute maximum. And somehow show that R = r.
To find the first derivative, I used the chain rule and came up with this:
f(g(x)) -> f'(g(x))g'(x)
P = RV^2/(R+r)^2 f = RV^2/x^2 g = R+r
f' = (2Vx - 2xRV^2)/x^4 g' = 0
P'= (2V(R+r) - 2(R+r)RV^2)/(R+r)^4
= (2V - 2RV^2)/(R+r)^3
(2V - 2RV^2)/(R+r)^3 = 0
I don't know if this is correct and I don't know how to go on from here. The denominator can't be zero because the derivative won't exist at those points. But if I set the nominator to zero, it doesn't really help me because there's no r, only R. What am I doing wrong?
All help is appreciated, thanks!