Question

The sales \(S\) (in billions of dollars per year) for Procter \& Gamble for the years 2001 through 2006 can be modeled by \(S=1.09312 t^{2}-1.8682 t+39.831, \quad 1 \leq t \leq 6\) where \(t\) represents the year, with \(t=1\) corresponding to 2001\. (Source: Procter \& Gamble Company) (a) During which year, from 2001 through 2006 , were Procter \& Gamble's sales increasing most rapidly? (b) During which year were the sales increasing at the lowest rate? (c) Find the rate of increase or decrease for each year in parts (a) and (b). (d) Use a graphing utility to graph the sales function. Then use the zoom and trace features to confirm the results in parts (a), (b), and (c).

Short answer

(a) The sales were increasing most rapidly in 2001. (b) The sales were increasing at the lowest rate in 2006. (c) The rate of increase for 2001 was approximately 0.32 billion dollars per year, and for 2006 was approximately 11.16 billion dollars per year.

Step-by-step solution

Find the Derivative of the Sales Function

The derivative of the function \(S(t) = 1.09312 t^{2}-1.8682 t+39.831\) with respect to \(t\) is given by \[S'(t) = 2 \cdot 1.09312 t - 1.8682\]. This derivative function tells us the rate of change of the sales (i.e., how fast the sales are increasing or decreasing) at any given year \(t\).

Determine the Sales Peak and Trough

To determine when the sales were increasing most rapidly and when they were increasing at the lowest rate, we need to find the critical points of the derivative function. The critical points can be found by setting \(S'(t)\) equal to zero and solving for \(t\), i.e., \(2 \cdot 1.09312 t - 1.8682 = 0\). Solving this equation yields \(t \approx 0.857\). This is the year when the sales trend changes from increasing at an increasing rate to increasing at a decreasing rate. However, because this does not fall within the given limit of \(t\) (2001 to 2006), we can conclude that the sales were increasing most rapidly in 2001 (t=1) and were increasing at the lowest rate in 2006 (t=6).

Calculate the Rate of Increase or Decrease

To find the rate of increase or decrease for the years determined in Step 2, we substitute \(t = 1\) and \(t = 6\) into \(S'(t)\). For \(t = 1\), we have \(S'(1) = 2 \cdot 1.09312 \cdot 1 - 1.8682 \approx 0.32\) billion dollars per year. And for \(t = 6\), we have \(S'(6) = 2 \cdot 1.09312 \cdot 6 - 1.8682 \approx 11.16\) billion dollars per year. So, the sales were increasing most rapidly at about 0.32 billion dollars per year in 2001, and were increasing at the lowest rate of about 11.16 billion dollars per year in 2006.

Key concepts

Derivative of Sales Function

Understanding the derivative of a sales function is crucial for businesses to measure growth and performance. In algebraic terms, the derivative represents the rate of change of the sales with respect to time. For the Procter & Gamble case, the function representing sales, S, over time, t, was given, and the derivative was found using basic calculus rules.

When we calculate the derivative, denoted as S'(t), we get an equation that provides the rate of change at any specific year. In the given period, by plugging in the values of t within the range of 1 to 6, it's possible to evaluate how fast (or slow) sales were increasing for each year. For instance, at t=1, the derivatives estimate showed the sales increase was about 0.32 billion dollars per year.

Critical Points in Calculus

Critical points in calculus are those where the derivative of a function is zero or undefined. They usually signify a change in the behavior of the function, such as switching from increasing to decreasing, or reaching a maximum or minimum value.

In the context of sales for a company, identifying critical points can hint at shifts in market dynamics or business performance. For Procter & Gamble's sales function, solving for when S'(t) = 0 should have provided the critical points. However, the critical value found did not lie within the years in question, showing the sales increased throughout the period but at varying rates. This highlights that a critical point may not always fall within the relevant domain, and context-specific interpretation is essential.

Rate of Change

The rate of change is a concept that demonstrates how a quantity alters over time. In sales analysis, it's a key indicator of business health, showing whether sales are growing, stagnating, or declining. By applying calculus and taking the derivative of the sales function, Procter & Gamble could determine the rate of change for sales on a year-by-year basis.

Specifically, a positive rate of change, as seen with S'(1), signifies an increase in sales, while a negative one would indicate a decline. The greater the value of this derivative, the more rapid the increase or decrease. The exercise provided illuminated the initial and final rates of increase, offering valuable insight into the sales trends over the defined period.

Graphing Utility in Mathematics

Graphing utilities are incredibly valuable in visualizing mathematical concepts and functions. In the case of Procter & Gamble, graphing the sales function allows for a visual representation of sales over time. By graphing the function S and its derivative S'(t), the rise and fall of sales, the slope, and curvature can be observed directly.

Using the zoom and trace features enables students to pinpoint rates of increase and confirm calculated results. Through graphical analysis, students can visually verify that sales were indeed increasing continually, complementing the algebraic findings, and in this way, graphing functions serve as an excellent tool for reinforcing understanding of calculus concepts.