Problem 36 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

It is estimated that between the hours of noon and 7:00 P.M., the speed of highway traffic flowing past a certain downtown exit is approximately $$ S(t)=t^{3}-9 t^{2}+15 t+45 $$ miles per hour, where \(t\) is the number of hours past noon. At what time between noon and 7:00 P.M. is the traffic moving the fastest, and at what time between noon and 7:00 P.M. is it moving the slowest?

Step-by-Step Solution

Answer

The traffic is moving the fastest at 1:00 P.M. and the slowest at 7:00 P.M.

Step 1: Identify the given function

The speed of highway traffic is given by the function S(t)=t^{3}-9t^{2}+15t+45 where t is the number of hours past noon.

Step 2: Find the derivative of the function

To find when the speed is fastest or slowest, determine the critical points by finding the derivative S'(t) . Calculate S'(t) : S'(t) = 3t^2 - 18t + 15

Step 3: Set the derivative equal to zero

To find the critical points, solve for t when S'(t) = 0 : 3t^2 - 18t + 15 = 0. Divide the entire equation by 3: t^2 - 6t + 5 = 0. Now factor the quadratic equation: (t - 1)(t - 5) = 0. This gives us t = 1 and t = 5 .

Step 4: Evaluate the function at critical points and endpoints

Determine the speed values at the critical points t = 1 and t = 5 , as well as at the endpoints t = 0 (noon) and t = 7 (7:00 P.M.). Calculate S(t) : S(0) = 0^3 - 9(0)^2 + 15(0) + 45 = 45 S(1) = 1^3 - 9(1)^2 + 15(1) + 45 = 52 S(5) = 5^3 - 9(5)^2 + 15(5) + 45 = -40 S(7) = 7^3 - 9(7)^2 + 15(7) + 45 = -53.

Step 5: Determine maximum and minimum values

The highest value from the calculations is S(1) = 52 , and the lowest value is S(7) = -53 . Therefore, the traffic is moving the fastest at t = 1 hour past noon (1:00 P.M.) and the slowest at t = 7 hours past noon (7:00 P.M.).

Key Concepts

critical points

Critical points are essential for understanding where a function reaches its highest and lowest values. To find these critical points, we first need to compute the derivative of the function, since critical points occur where this derivative is zero or undefined.
In our problem, the speed function given is
\[ S(t) = t^3 - 9t^2 + 15t + 45 \]
After computing the derivative, we set it to zero to solve for the values of \( t \).
These values represent potential times at which the speed of the highway traffic might be at a maximum or minimum.
For our function, the derivative is:
\( S'(t) = 3t^2 - 18t + 15 \)
Setting this equal to zero gives:
\[ 3t^2 - 18t + 15 = 0 \]
These are the critical points we need to further investigate.

derivative

The derivative of a function provides the rate at which the function's value is changing. It's a tool we use to determine where the function's slope is zero, meaning either a peak, trough, or a point of inflection.
In our specific problem, the speed function \( S(t) \) is:
\[ S(t) = t^3 - 9t^2 + 15t + 45 \]
The derivative of this function—\( S'(t) \)—is found using power rules of differentiation:
\[ S'(t) = 3t^2 - 18t + 15 \]

By calculating this, we've now set the stage for finding where speeds are extremal (fastest or slowest). The zeros of this derivative are the values of \( t \) where the speed might either be maximized or minimized.

maximum and minimum values

Maximum and minimum values tell us the highest and lowest points of a function over a certain interval. To find them, evaluate the function at critical points and endpoints.
In our scenario, we found critical points by solving \( 3t^2 - 18t + 15 = 0 \) and factored it down to:
\[ (t - 1)(t - 5) = 0 \]
Therefore, \( t = 1 \) and \( t = 5 \) are critical points. We also consider endpoints \( t = 0 \) and \( t = 7 \).
Evaluating the function at these points, we get:
\[ S(0) = 45 \]
\[ S(1) = 52 \]
\[ S(5) = -40 \]
\[ S(7) = -53 \]
The maximum value of 52 occurs at 1 hour past noon, and the minimum value of -53 occurs at 7 hours past noon.

quadratic equation factoring

Quadratic equation factoring is a critical skill for solving problems in calculus and algebra. Factoring allows us to break down equations into simpler forms, making them easier to solve.
When finding critical points, we factored the quadratic equation \( 3t^2 - 18t + 15 \) to simplify it:
First, we divided the entire equation by 3 to get:
\[ t^2 - 6t + 5 = 0 \]
Then, we factored this into:
\[ (t - 1)(t - 5) = 0 \]
The solutions, \( t = 1 \) and \( t = 5 \), were obtained through factoring. This was crucial in determining where the function changes direction and ultimately helped in identifying maximum and minimum values.