When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance, R, can be calculated from the equation 1R=1r1+1r2, where R, r1, and r2 are positive. Assume that r2 is constant.

(a) Show that R is an increasing function of r1.

(b) Where on the interval a≤r1≤b does R take its maximum value?

a)

The combined resistance of a circuit consisting of two resistors in parallel is given by:

where

R is the combined resistance

are the two resistors

We can re-write the expression as follows:

Or

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

b)

To solve this part, we use again the expression for R written in part a:

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

So the interval of allowed values for r1 is

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a