Question

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) = 85s / 20 + 20( s / 23 )²
at what speed can the greatest number of cars travel safely on that road?

Explanation:

Step1: Differentiate the function

We have \(N(s)=\frac{85s}{20 + 20(\frac{s}{23})^2}\). First, simplify the denominator: \(D = 20+20(\frac{s^{2}}{529})=\frac{20\times529 + 20s^{2}}{529}\). So \(N(s)=\frac{85s\times529}{20(529 + s^{2})}=\frac{44965s}{20(s^{2}+529)}\).

Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 44965s\), \(u^\prime=44965\), \(v = 20(s^{2}+529)\), \(v^\prime = 40s\).

\(N^\prime(s)=\frac{44965\times20(s^{2}+529)-44965s\times40s}{400(s^{2}+529)^{2}}\)

Simplify the numerator:

$$
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$$

So \(N^\prime(s)=\frac{44965\times20(529 - s^{2})}{400(s^{2}+529)^{2}}\)

Step2: Find the critical points

Set \(N^\prime(s)=0\). Since the denominator \(400(s^{2}+529)^{2}>0\) for all real \(s\), we set the numerator equal to zero.

\(44965\times20(529 - s^{2}) = 0\)

\(529 - s^{2}=0\)

\(s^{2}=529\)

\(s=\pm23\). Since speed \(s>0\), we consider \(s = 23\)

Step3: Second - derivative test (optional, but can confirm maximum)

Differentiate \(N^\prime(s)\) again using the quotient rule. But since the function \(N(s)\) is a rational function and we know from the first - derivative that at \(s = 23\), the first - derivative changes sign from positive (for \(s<23\)) to negative (for \(s>23\))

Answer:

The speed at which the greatest number of cars can travel safely on the road is \(s = 23\) units of speed (assuming the units of \(s\) are consistent with the problem's context).