In a 120 volt electrical circuit having a resistance of 16 ohms, the available power P in watts is a function of I, the amount of current flowing in amperes. If P(I)=120I-16I^2, how many amperes will produce the maximum power in the circuit? What is the maximum power?

Answer 1

Calculus:
dP/dI=0=120-32I....I=120/32 amps.
Graphing:
y=-16x^2+120x =x(-16x+120) and those roots (when y=0) (x=0, or x=120/16)
well, it is a parabola, crossing the x axis at those roots, so the max should be half way between the roots, or at x=1/2 (120/16 +0) + O = 120/32