Question

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

F= 3600v/(22+0.07v^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

about 17.7 mph

Explanation:

You want the value of v that maximizes F(v) = 3600v/(22+0.07v²).

Derivative

The derivative of F is ...

`F'(v)=(3600(22+0.07v^2)-3600v(0.14v))/((22+0.07v^2)^2)`

Maximum

The derivative is zero at the maximum. This is the case when the numerator is zero:

3600(22 +.07v² -.14v²) = 0

.07v² = 22

v = √(22/0.07) ≈ 17.728

Flow rate is maximized at a speed of about 17.7 miles per hour.