Problem 58 - Algebra and Trigonometry with Analytic Geometry

Question

Certain learning processes may be illustrated by the graph of an equation of the form \(f(x)=a+b\left(1-e^{-c}\right)\), where \(a, b\), and \(c\) are positive constants. Suppose a manufacturer estimates that a new employee can produce five items the first day on the job. As the employee becomes more proficient, the daily production increases until a certain maximum production is reached. Suppose that on the \(n\)th day on the job, the number \(f(n)\) of items produced is approximated by $$ f(n)=3+20\left(1-e^{-0.1 n}\right) . $$ (a) Estimate the number of items produced on the fifth day, the ninth day, the twenty-fourth day, and the thirtieth day. (b) Sketch the graph of \(f\) from \(n=0\) to \(n=30\). (Graphs of this type are called learning curves and are used frequently in education and psychology.) (c) What happens as \(n\) increases without bound?

Step-by-Step Solution

Answer

On day 5, produce 11 items; day 9, 15 items; day 24, 22 items; day 30, 23 items. The graph shows rapid early growth, leveling at maximum 23 items.

Step 1: Calculate Items Produced on Specific Days

Substitute every given day into the function \( f(n) = 3 + 20(1 - e^{-0.1n}) \) and calculate the result. For the fifth day:\[ f(5) = 3 + 20(1 - e^{-0.5}) \]. Use a calculator to find \( e^{-0.5} \approx 0.6065 \), then:\[ f(5) = 3 + 20(1 - 0.6065) \approx 10.87 \] (rounding to the nearest whole number, \( 11 \) items).For the ninth day:\[ f(9) = 3 + 20(1 - e^{-0.9}) \].Using \( e^{-0.9} \approx 0.4066 \), then:\[ f(9) = 3 + 20(1 - 0.4066) \approx 14.87 \] (rounding, \( 15 \) items).For the twenty-fourth day:\[ f(24) = 3 + 20(1 - e^{-2.4}) \].Evaluate \( e^{-2.4} \approx 0.0907 \), therefore:\[ f(24) = 3 + 20(1 - 0.0907) \approx 21.814 \] (rounding, \( 22 \) items).For the thirtieth day:\[ f(30) = 3 + 20(1 - e^{-3}) \].With \( e^{-3} \approx 0.0498 \), we find:\[ f(30) = 3 + 20(1 - 0.0498) \approx 22.996 \] (rounding, \( 23 \) items).

Step 2: Sketch the Graph

Plot the function \( f(n) = 3 + 20(1 - e^{-0.1n}) \) for values from \( n = 0 \) to \( n = 30 \). The curve starts at \( n=0 \) with an approximate production of 3 items and rises up quickly at first, then slows around \( n=15 \), gradually nearing a horizontal asymptote as it approaches closer to the maximum production capacity as \( n \) increases. The graph shows the traditional 'S' shape common for learning curves, illustrating rapid early learning followed by a leveling out.

Step 3: Determine Asymptotic Behavior

As \( n \) increases towards infinity, the term \( e^{-0.1n} \) gets closer to 0. Therefore, the function \( f(n) = 3 + 20(1 - e^{-0.1n}) \) approaches \( f(n) = 3 + 20 = 23 \). Thus, the production of items levels off, meaning the employee cannot produce more than 23 items even as they continue to work.

Key Concepts

Exponential Growth

Exponential growth is a powerful phenomenon where quantities increase by a fixed percentage over equal increments of time. In our exercise, the learning curve is described by the function \( f(n) = 3 + 20(1 - e^{-0.1n}) \). The term \( e^{-0.1n} \) is an example of exponential decay, rapidly decreasing as \( n \) increases. This fast-decreasing term creates the initial rapid growth observed in our learning curve. As the exponential component approaches zero, the growth rate of \( f(n) \) slows down. This behavior mirrors how individuals or systems experience accelerated learning initially, followed by slower improvements as they gain proficiency.

Asymptotic Behavior

Understanding asymptotic behavior involves analyzing what happens to a function as the independent variable approaches infinity. For the learning curve function \( f(n) = 3 + 20(1 - e^{-0.1n}) \), as \( n \) becomes very large, \( e^{-0.1n} \) trends towards zero. This makes \( f(n) \) approach the value \( 3 + 20 \), or \( 23 \). At this point, the learning curve reaches its asymptote, indicating maximum production capacity. The asymptotic behavior reflects the practical limits of a process, revealing that beyond a certain point, additional practice yields minimal gains in performance.

Graph Sketching

Sketching a graph provides a visual representation of how a function behaves over a range of values. For our function, graphing \( f(n) = 3 + 20(1 - e^{-0.1n}) \) from \( n = 0 \) to \( n = 30 \) illustrates an S-shaped learning curve common in growth processes. Begin plotting points by calculating \( f(n) \) for integers like 0, 5, 10, 20, and 30. Initially steep, the curve flattens as \( n \) increases, reflecting the diminishing returns in learning. The graph can show both rapid early growth and eventual stabilization, aiding in understanding the transition from novice to proficient.

Production Estimation

Production estimation involves using mathematical models to forecast outcomes over time. For the given scenario, the function \( f(n) = 3 + 20(1 - e^{-0.1n}) \) estimates production potential as a worker learns. Calculating \( f(n) \) for specific days can predict items produced—the given examples show 11 items on the 5th day, 15 on the 9th, and stabilizing around 23 items by the 24th and 30th days. Such estimations guide decision-making in organizational settings, planning for staff training periods and resource allocation by predicting when workers will reach peak productivity.