Question

On a given day, the flow rate F (cars per hour) on a congested roadway is

F= v/(22+0.02v^2)

where v is the speed of the traffic in miles per hour. What speed will maximize the flow rate on the road?

Answer 1

have you taken the derivative yet?

Answer 2

not yet

Answer 3

see if you can find it, then set it equal to zero and solve vor v. That will give you the location of any extrema for F.

Answer 4

if you're having trouble finding it let me know.

Answer 5

can you give to me by step?

Answer 6

use the quotient rule:
\[(\frac{ U }{V })' = \frac{ U ' V- UV ' }{ V ^{2}}\]

Answer 7

an then?

Answer 8

find the derivative of the numerator and denominator and plug them into that expression where appropriate

Answer 9

derivative of the numerator is 1 so:\[\frac{ U'V -UV' }{ V ^{2} } = \frac{ 1*(22+.02v ^{2)} - UV'}{ V ^{2} }\]

Answer 10

continue... find the derivative of the denominator...

Answer 11

what will it be?

Answer 12

(22+0.02v^2)^2

Answer 13

\[\frac{ U'V -UV' }{ V ^{2} } = \frac{ 1*(22+.02v ^{2}) - v*.04v}{ V ^{2} }\]

Answer 14

then i will get also the 2nd derivative?

Answer 15

\[\frac{ U'V -UV' }{ V ^{2} } = \frac{ 1*(22+.02v ^{2})-v*.04v }{ (22+.02v ^{2})^{2} }\]

Answer 16

what is the next stepif i already knew the 1st derivative?

Answer 17

2nd derivative?

Answer 18

no, set the derivative equal to zero and solve for v

Answer 19

\[f'(x)=(22+.02v2)-.04v^2 / (22+0.02v^2)^2\]

Answer 20

pretty much.. now set it equal to zero and solve for v

Answer 21

\[(22+.02v^2)-.04v^2/(22+0.02v^2)=0\]