Revenue decreasing per day

[geology]

I'm stumped on this one. Any help would be appreciated.

The revenue from parking fines in a town is given by R(n)=(8000n)/(n+2) dollars, where n is the number of parking patrol hours per day. At the outbreak of a flu epidemic, 30 patrol hours were being used daily. During the epidemic, this number was decreasing at a rate of 6 patrol hours per day. How fast is the parking revenue decreasing per day, at the outbreak of the flu epidemic?

a. $93.75 per day
b. $15.63 per day
c. $1000 per day
d. $142.01 per day
e. $23.67 per day

I am not sure where to take this problem. On one attempt I differentiated (isn't this marginal revenue? It didn't seem right but I tried it anyway) R'(n)= 16000/(x+2)^2 and then inserted what I believed to be n = 30-6d coming up with $23.67 dollars per day. This is one of the answers in the multiple choice, but the answer key says that the answer should be $93.75 per day. I also attempted to use the formula: R(30) - R(24) = Change in Revenue per day. This way didn't seem to work either. Can you please tell me what steps are needed to solve this equation.

 

[tkhunny]

\(\displaystyle dR = \frac{16000}{(n+2)^{2}}dn\)

\(\displaystyle dR = \frac{16000}{(30+2)^{2}}(-6)\)

I agree with the book. I can't tell what it is you are doing.

 

[geology]

Well, it looks like I was on the right track taking the derivative, my first attempt, but I didnt multiply by the derivative of n. I'm teaching myself calculus out of a book unfortunately and I am a little lost as to why you multiply the derivative of R by the derivative of n to get the final product? I'm sure this is something simple, but I am just stumped as to why you do that sometimes, but not others. Is this a specific kind of differentiation?

 

[tkhunny]

It's rather a trick of the notation. The derivative notation is more useful than you might think.

You wrote R'(n). This is "Function Notation" and sugests that the derivative is a function.

You may wrote \(\displaystyle \frac{dR}{dn}\) to mean essentially the same thing, "The derivative of a function, "R", with respect to the independent variable 'n'." or just "The derivative or R wrt n". This is also called the "differential form.

The fun part is that the differential form can be broken up to emphasize that R changes with a predictable relationship to n. This is what happened when I left dR on one side and moved dn to the other.

 

[Deleted member 4993]

\(\displaystyle \frac{dR}{dt} \ = \ \frac{dR}{dn}\cdot \frac{dn}{dt}\)

 

[geology]

\(\displaystyle \frac{dR}{dt} \ = \ \frac{dR}{dn}\cdot \frac{dn}{dt}\)

I think I get it: It is because we are looking for the derivative of Revenue with respect to time ("How fast is the revenue decreasing per day...") but we are essentially solving with a given amount of days ("beginning of the outbreak..." = 1 day) and the given number of patrol hours (30). Sort of like cross canceling in a chemical reaction?

 

[mmm4444bot]

R(d) is a function within a function. (Google "composite functions").

R is a function of n, but n itself is a function of d.

You can think of the function for n in terms of d as: n(d) = 30 - 6d.

First, some number d goes into function n, and the number n(d) comes out.

Next, this output number n(d) goes into function R, and the number R(n(d)) comes out.

Note: Subhotosh changed the symbol name from d to t, in his example. I will too, in what follows.

dR/dt is "the rate-of-change in R with respect to t"

We see that this rate is a product of two other rates:

dR/dn is "the rate-of-change in R with respect to n" and dn/dt is "the rate-of-change of n with respect to t".

In short, if the number t (which is called d in your exercise) changes, that change affects the rate at which R changes, according to the Chain Rule in calculus.