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If a resistor of $ R $ ohms is connected across a battery of $ E $ volts with internal resistance $ r $ ohms, then the power (in watts) in the external resistor is$$ P = \dfrac{E^2 R}{(R + r)^2} $$If $ E $ and $ r $ are fixed but $ R $ varies, what is the maximum value of the power?

$\frac{E^{2}}{4 r}$

02:02

Wen Z.

00:54

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Baylor University

University of Michigan - Ann Arbor

Boston College

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we're told a resistor of big are owns is connected across the battery of big ive bolts with internal resistance. Little are owns and the power in Watts and the external resistor is given by the equation. Big P equals big e squared times big are over big r plus Little are squared. Mhm. Now if Big E and little are are fixed but big are varies were asked what the maximum value of our power is. First we'll find the derivative treating biggie and little ours constants and bigger is the variable. So we have a ratio. You have that p prime of big are is the bottom big R plus little r squared times the derivative the top e squared minus the top e squared big are times that are at the bottom which is to bigger Plus little are all over the bottom Bigger plus little r squared squared, which is big are possible Large Fourth, this simplifies Eventually Two big e squared little ar minus big e squared big are over big r plus little are cute So I skipped a few steps here. You might want to do these on your own. We want to find when this is equal to zero to find our critical values. Well, clearly, this is only equal to zero when the numerator is zero. And so when big R is equal to little are now we find that when bigger is less than little are Well, then we have that are derivative keep prime of big are well, this is greater than zero And when big R is greater than little are p prime of big art is less than zero. Therefore, it follows that at the first derivative test he has a maximum value at a big are equals Little are plugging in big are equals Little are we find the maximum power p of little are this is e squared times little are over Yeah, little R plus little r squared, which is the same as you squared over four Little are

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