# Maximum power without calculus

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A Quora question asks: My physics textbook says that in a simple series circuit with one resistor R and a battery with internal resistance r the maximum power dissipated in R is when R equals the internal resistance r, why is this the case?

Mahesh Prakash is correct that this is a simple exercise in calculus (and in fact it may have been given as a homework problem in a course for students who have learned some calculus). But it may be useful for some to see if we can prove it without using calculus.

Given that the power dissipated in a resistor is given by current times voltage (analogous to the mechanical situation where power is speed times force), and that in a series circuit the current is equal to the total EMF divided by the total resistance, we find that our goal is to maximize P=IV=I^2R=(\frac{E}{R+r})^{2}R, where E (being the voltage across the battery when no current is flowing) and r (being the internal resistance of the battery) are fixed properties of just the battery, and our only control variable is R.

Now increasing a number decreases its reciprocal (and vice versa) and multiplying by a positive constant doesn’t change the locations of its max and min. So the max of P occurs at the same R value as the min of \frac{(R+r)^2}{R}=\frac{R^2+2Rr+r^2}{R}=R+2r+\frac{r^2}{R}=r(\frac{R}{r}+2+\frac{r}{R}).

This occurs for the same R as the min of \frac{R}{r}+\frac{r}{R}= q+\frac{1}{q} with q=\frac{R}{r}, and our goal is to show that this happens when q=1.

Clearly, for positive x, y=x+\frac{1}{x} gets large when x is either very large or very small, but how can we see that it bottoms out at x=1?

Well, one way is to check that it is always going down to the right for x<1 and up for x>1.

(That’s easy using calculus but our goal here is to show it for someone who has not yet studied calculus.)

If we consider the effect of increasing x a bit, to say x_{+}, then the change of y is (x_{+}+\frac{1}{x_{+}})-(x+\frac{1}{x})=(x_{+}-x)+(\frac{1}{x_{+}}-\frac{1}{x}).

When the first of these is positive the other is negative and it’s a matter of checking which one wins.

But |(\frac{1}{x_{+}}-\frac{1}{x})|=\frac{x_{+}-x}{x_{+}x}, and when x and x_{+} are both bigger than 1 it is less than x_{+}-x, so the increase of x beats the decease of 1/x. And when x and x_{+} are both less than 1 it goes the other way and the decease of 1/x beats the increase of x.

# Why SD uses squares (rather than abs val)

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It’s not just for computational convenience.

rms SD minimizes expected distance from mean while avg of abs val does it for the median