Question

optimization: problem 3
(1 point)
a traffic fanatic wants to estimate the maximum number of cars that can safely travel on a particular road at a given speed. he assumes that each car is 19 feet long, travels at speed s, and follows the car in front of it at a safe distance for that speed. he finds that the number n of cars that can pass a given spot per minute is modeled by the function
n(s)=\frac{82s}{19 + 19(\frac{s}{22})^2}
at what speed can the greatest number of cars travel safely on that road?
s = \square

Explanation:

Step1: Simplify the function

First, simplify \(N(s)=\frac{82s}{19 + 19(\frac{s}{22})^2}\). Let \(y = N(s)\), then \(y=\frac{82s}{19+\frac{19s^{2}}{484}}=\frac{82s\times484}{19\times484 + 19s^{2}}=\frac{484\times82s}{19(s^{2}+484)}\). Since \(\frac{484\times82}{19}\) is a constant, we can focus on maximizing the function \(f(s)=\frac{s}{s^{2}+484}\) (by the property of proportionality \(y = k\times f(s)\) where \(k=\frac{484\times82}{19}\)).

Step2: Use the quotient - rule for differentiation

The quotient rule states that if \(y=\frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). For \(f(s)=\frac{s}{s^{2}+484}\), let \(u = s\), \(u^\prime=1\), \(v=s^{2}+484\), \(v^\prime = 2s\). Then \(f^\prime(s)=\frac{(s^{2}+484)-s\times(2s)}{(s^{2}+484)^{2}}=\frac{s^{2}+484 - 2s^{2}}{(s^{2}+484)^{2}}=\frac{484 - s^{2}}{(s^{2}+484)^{2}}\).

Step3: Find the critical points

Set \(f^\prime(s)=0\). Since the denominator \((s^{2}+484)^{2}>0\) for all real \(s\), we set the numerator equal to zero. So \(484 - s^{2}=0\), which gives \(s^{2}=484\). Solving for \(s\) (and considering \(s>0\) because speed cannot be negative), we have \(s = 22\).

Step4: Use the second - derivative test (or analyze the sign of the first - derivative)

For \(s\in(0,22)\), let \(s = 10\), then \(f^\prime(10)=\frac{484-100}{(100 + 484)^{2}}=\frac{384}{(584)^{2}}>0\). For \(s\in(22,\infty)\), let \(s = 30\), then \(f^\prime(30)=\frac{484 - 900}{(900+484)^{2}}=\frac{-416}{(1384)^{2}}<0\). So the function \(f(s)\) (and thus \(N(s)\)) has a maximum at \(s = 22\).

Answer:

\(22\)