Question

optimization: problem 3
(1 point)
a highway engineer wants to estimate the maximum number of cars that can safely travel on a particular road at a given speed. she assumes that each car is 20 feet long, travels at speed s, and follows the car in front of
it at a safe distance for that speed. she finds that the number n of cars that can pass a given spot per minute is modeled by the function
n(s) = \frac{85s}{20 + 20(\frac{s}{22})^2}
at what speed can the greatest number of cars travel safely on that road?
s = \square
submit answers

Explanation:

Step1: Find the derivative of \(N(s)\)

We have \(N(s)=\frac{85s}{20 + 20(\frac{s}{22})^2}\). First, rewrite it as \(N(s)=\frac{85s}{20+\frac{20s^{2}}{484}}=\frac{85s\times484}{20\times484 + 20s^{2}}=\frac{41140s}{9680+20s^{2}}\).
Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 41140s\), \(u^\prime=41140\), \(v = 9680 + 20s^{2}\), \(v^\prime = 40s\).
\(N^\prime(s)=\frac{41140(9680 + 20s^{2})-41140s\times40s}{(9680 + 20s^{2})^{2}}=\frac{41140\times9680+41140\times20s^{2}-41140\times40s^{2}}{(9680 + 20s^{2})^{2}}=\frac{41140\times9680-41140\times20s^{2}}{(9680 + 20s^{2})^{2}}\).
Set \(N^\prime(s)=0\), then \(41140\times9680-41140\times20s^{2}=0\).
Divide both sides by \(41140\): \(9680-20s^{2}=0\).

Step2: Solve for \(s\)

From \(9680-20s^{2}=0\), we get \(20s^{2}=9680\), \(s^{2}=\frac{9680}{20}=484\).
Take the square - root: \(s=\pm22\). Since speed \(s>0\), we take \(s = 22\).

Answer:

\(22\)