Question

(b) suppose that the manager of the food emporium wanted to use the average rate of change of the number of customers in the store over the interval from 6:00 a.m. to 7:30 a.m. on this day to predict the number of customers that were in the store at 9:00 a.m. without actually estimating the number of customers in the store, explain whether using the average rate of change over the interval from 6:00 a.m. to 7:30 a.m. would yield an overestimate or an underestimate of the actual number of customers in the store. provide a justification for your answer.
(c) the manager did not record the customer count at 2:00 p.m. on this day. use the average rate of change between 1:30 p.m. and 3:30 p.m. to estimate the number of customers that were in the store at 2:00 p.m. on this day.
(d) the manager of the store is thinking about adding another employee to a six - hour shift either from 6:00 a.m. to 12:00 p.m. or from 12:00 p.m. to 6:00 p.m. during which time interval would you recommend that the manager add this employee? provide an explanation that includes a discussion of the average rates of change over the intervals indicated.
(e) the assistant manager notices that the number of customers in the store at 5:30 p.m. is equal to the number of customers in the store at 5:00 p.m. the assistant manager concludes that because the average rate of change is zero, then no customers left or entered between 5:00 p.m. and 5:30 p.m. is the assistant managers reasoning correct? explain why or why not.

Explanation:

Step1: Analyze the average rate of change concept

The average rate of change formula is \(\frac{f(b)-f(a)}{b - a}\). If the function (number of customers over time) is concave - up (increasing at an increasing rate) or concave - down (increasing at a decreasing rate), using the average rate of change over a larger interval to estimate a value within a sub - interval can lead to over - or under - estimation.

Step2: Determine the concavity for the 6:00 a.m. - 9:00 a.m. to 7:30 a.m. case

Assume the number of customers is a function \(y = f(t)\) of time \(t\). From 6:00 a.m. to 9:00 a.m., if the rate at which customers enter the store is increasing (the function is concave - up), then the average rate of change over the larger interval \([6,9]\) is less than the average rate of change over the sub - interval \([6,7.5]\). Using the average rate of change over \([6,9]\) to estimate the number of customers at 7:30 a.m. (which is in the sub - interval \([6,9]\)) will under - estimate. Because the slope (rate of change) of the secant line over \([6,9]\) is less than the slope of the secant line over \([6,7.5]\) if the function is concave - up (customers are arriving at an increasing rate).

Step3: Analyze the 1:30 p.m. - 3:30 p.m. average rate of change for estimating 2:00 p.m.

Let \(t_1 = 1.5\) (1:30 p.m. in terms of hours after 12:00 p.m.), \(t_2=3.5\) (3:30 p.m. in terms of hours after 12:00 p.m.). The average rate of change formula \(\frac{f(t_2)-f(t_1)}{t_2 - t_1}\). If we assume \(f(t)\) is a linear function (a rough estimate using the average rate of change), and we want to find \(f(2)\) (2:00 p.m. or \(t = 2\) in terms of hours after 12:00 p.m.). We know \(f(2)=f(1.5)+(2 - 1.5)\times\frac{f(3.5)-f(1.5)}{3.5 - 1.5}\)

Step4: Analyze the average rate of change for the employee - adding problem

The average rate of change \(\frac{\Delta y}{\Delta t}=\frac{y_2 - y_1}{t_2 - t_1}\). Calculate the average rate of change for the 6:00 a.m. - 12:00 p.m. interval and the 12:00 p.m. - 6:00 p.m. interval. If the average rate of change (number of customers per hour) is higher in one interval, that interval is busier. For example, if the number of customers at 6:00 a.m. is \(y_1\), at 12:00 p.m. is \(y_2\), and at 6:00 p.m. is \(y_3\). The average rate of change for 6:00 a.m. - 12:00 p.m. is \(\frac{y_2 - y_1}{12 - 6}\), and for 12:00 p.m. - 6:00 p.m. is \(\frac{y_3 - y_2}{6 - 0}\). If \(\frac{y_2 - y_1}{6}>\frac{y_3 - y_2}{6}\), then the 6:00 a.m. - 12:00 p.m. interval is busier.

Step5: Analyze the zero - average - rate - of - change reasoning

The average rate of change formula \(\frac{f(5.5)-f(5)}{5.5 - 5}\). If \(\frac{f(5.5)-f(5)}{0.5}=0\), then \(f(5.5)=f(5)\). But this only means that the net change in the number of customers is zero. Customers could have left and entered the store such that the number of customers at 5:00 p.m. (\(f(5)\)) and 5:30 p.m. (\(f(5.5)\)) is the same. For example, if \(x\) customers left and \(x\) customers entered in the 30 - minute interval.

Answer:

(b) Using the average rate of change over the interval from 6:00 a.m. to 9:00 a.m. to estimate the number of customers at 7:30 a.m. will under - estimate. Justification: If the function representing the number of customers over time is concave - up (customers are arriving at an increasing rate), the average rate of change over the larger interval \([6,9]\) is less than the average rate of change over the sub - interval \([6,7.5]\).
(c) Let \(t_1 = 1.5\) (1:30 p.m.), \(t_2 = 3.5\) (3:30 p.m.). If \(f(t)\) is the number of customers as a function of \(t\) (hours after 12:00 p.m.), then \(f(2)=f(1.5)+(2 - 1.5)\times\frac{f(3.5)-f(1.5)}{3.5 - 1.5}\) (use the values of \(f(1.5)\) and \(f(3.5)\) from the data table to calculate the numerical estimate).
(d) Calculate the average rate of change for 6:00 a.m. - 12:00 p.m. (\(\frac{y_{12}-y_{6}}{6}\)) and 12:00 p.m. - 6:00 p.m. (\(\frac{y_{18}-y_{12}}{6}\)). Recommend the interval with the higher average rate of change. For example, if \(\frac{y_{12}-y_{6}}{6}>\frac{y_{18}-y_{12}}{6}\), recommend 6:00 a.m. - 12:00 p.m. because a higher average rate of change means more customers per hour (more business).
(e) The assistant manager's reasoning is not correct. A zero average rate of change \(\frac{f(5.5)-f(5)}{0.5}=0\) (where \(f(t)\) is the number of customers at time \(t\)) only implies \(f(5.5)=f(5)\), but customers could have left and entered the store in the 30 - minute interval such that the net change is zero.