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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?

Verified Answer :

Answer:

33.16 mph

EXPLANATION:

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

set F'=0 and solve to get the square root of 1100, which is about 33.16 mph.

The other person's answer is wrong :(, trust me I've looked it up on several other places as well.

Cricetus

The speed that will maximize the flow rate on the road is "1.13 miles/hour".

According to the question,

→ \(F = \frac{v}{(22+0.02 \ v^2)}\)

We need to take the directive of the function as well as equation it to zero, then

→  \(F = \frac{v}{22} +0.02 \ v^2\)

→ \(F' = \frac{1}{22} +0.04 v\)

When \(F' =0\),

→ \(\frac{1}{22} +0.04 v =0\)

→         \(0.04v = -0.0454\)

→               \(v = -\frac{0.0454}{0.04}\)

→                  \(= -1.13\)

Thus the above answer answer i.e., 1.13 miles/hour is correct.

Learn more:

https://brainly.com/question/2195736

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