Question

On a given day, the flow rate F (cars per hour) on a congested roadway is F =v / (22 + .02v^2) where v is the speed in miles per hour. What speed will maximize the flow rate on the road?

Answer 1

Well, to optimize, you would normally need to find 2 related equations, substitute, derivative set to 0, back substitute. But you already have an equation give to you, so we can skip striaght to the derivative part.
So if you want to find a min or a max, you just need to take the derivative your equation and then set the NUMERATOR equal to 0 then solve for v. Think you cna get the derivative?

Answer 2

Ah, okay. Yeah, gimme a minute. :D

Answer 3

kk, np

Answer 4

Agh, I'm getting \[\frac{ v \times .04v }{ 22+.02v ^{2} }\] as the derivative and I feel like that is incorrect. ><;

Answer 5

Should be this:

\[\frac{ 22-.02v ^{2} }{ (22+.02v ^{2})^{2} }\]
Think you can maybe see how I gotthere?

Answer 6

Well I used \[\frac{ f' g - fg'}{ g ^{2}} \] to get \[ \frac{ 1 \times (22 + .02v ^{2}) - v \times .04v }{ 22 + .02v ^{2} }\] (sorry meant to square the bottom) and canceled out the common.

Answer 7

Yeah, there ya go. So now set the numerator equal to 0. We don't do the denominator equal to 0 because these only give us undefined values and undefined slopes, etc.

Answer 8

I'm getting v = 33.1662

Answer 9

That's what I get : )

Answer 10

Awesome! Thanks for the help!

Answer 11

yep yep : )