Problem 46 - Applied Calculus

Question

(a) Between 1995 and 2005, ACME Widgets sold widgets at a continuous rate of \(R=R_{0} e^{0.15 t}\) widgets per year, where \(t\) is time in years since January 1 . 1995\. Suppose they were selling widgets at a rate of 1000 per year on January 1, 1995. How many widgets did they sell between 1995 and 2005 ? How many did they sell if the rate on January 1,1995 was \(150,000,000\) widgets per year? (b) In the first case (1000 widgets per year on January 1,1995 ), how long did it take for half the widgets in the ten-year period to be sold? In the second case \((150,000,000\) widgets per year on January 1,1995 ), when had half the widgets in the ten-year period been sold? (c) In 2005, ACME advertised that half the widgets it had sold in the previous ten years were still in use. Based on your answer to part (b), how long must a widget last in order to justify this claim?

Step-by-Step Solution

Answer

ACME sold 14,298 widgets (initial rate: 1000/year) and 2.14 billion widgets (initial rate: 150 million/year) in 10 years. Half of these were sold after approximately 4.62 years (1000/year) and 5.38 years (150 million/year), needing the widgets to last around 5 years.

Step 1: Understanding the Sales Rate Function

The sales rate function is given as \(R = R_0 e^{0.15t}\), where \(t\) is the time in years since January 1, 1995. The initial sales rate, \(R_0\), changes between two scenarios: 1000 widgets per year and 150,000,000 widgets per year.

Step 2: Integrating to Find Total Widgets Sold

To find the total number of widgets sold over the ten-year period from 1995 to 2005, integrate the sales rate function from \(t = 0\) to \(t = 10\). This gives:\[\int_0^{10} R_0 e^{0.15t} \, dt = R_0 \left[ \frac{e^{0.15t}}{0.15} \right]_0^{10}.\]

Step 3: Calculating Widgets Sold for 1000 Widgets/Year Initial Rate

Substitute \(R_0 = 1000\) into the integral:\[1000 \left[ \frac{e^{0.15 \times 10}}{0.15} - \frac{e^{0.15 \times 0}}{0.15} \right] = 1000 \times \left[ \frac{e^{1.5} - 1}{0.15} \right].\]Calculate the expression to find the total widgets sold.

Step 4: Calculating Widgets Sold for 150,000,000 Widgets/Year Initial Rate

Substitute \(R_0 = 150,000,000\) into the integral:\[150,000,000 \left[ \frac{e^{1.5} - 1}{0.15} \right].\]Evaluate this expression for the total widgets sold.

Step 5: Determining When Half the Widgets Were Sold (1000 Widgets/Year Case)

Set up the equation to find when half the widgets were sold:\[\int_0^{t} 1000 e^{0.15t} \, dt = \frac{1}{2} \times 1000 \left[ \frac{e^{1.5} - 1}{0.15} \right].\]Solve this equation for \(t\) to determine the time when half were sold.

Step 6: Determining When Half the Widgets Were Sold (150,000,000 Widgets/Year Case)

Set up the equation to find when half the widgets were sold:\[\int_0^{t} 150,000,000 e^{0.15t} \, dt = \frac{1}{2} \times 150,000,000 \left[ \frac{e^{1.5} - 1}{0.15} \right].\]Solve this equation for \(t\) to determine the time when half were sold.

Step 7: Estimating Widget Lifespan

Based on your calculations in Step 5 and Step 6, estimate how long widgets remain in use to support the claim that half remained in use by 2005. Compare this lifespan to the earlier determined time when half were initially sold.

Key Concepts

Exponential Function

An exponential function is a mathematical expression in which a constant rate of change results in a rapid increase or decrease.
In the context of the continuous growth model, this function is defined as:

• The equation: \( R = R_0 e^{kt} \)
• Where \( R_0 \) is the initial rate, \( e \) is Euler's number (approximately 2.71828), \( k \) is the growth constant, and \( t \) is time.

This model is ideal for situations like continuously increasing sales rates, where the amount sold compounds over time at a constant percentage.
In our exercise, the growth constant \( k \) is 0.15, representing a 15% annual increase in sales. This function allows us to predict how many widgets will be sold at any given time during the ten-year span.
By understanding the structure of the exponential function, we can grasp how sales evolve and why they ramp up rapidly over the years.

Integration

Integration is a fundamental concept in calculus that allows us to determine the total accumulated amount over a period.
In this scenario, we're using integration to find the total number of widgets sold over ten years. We integrate the sales rate function from the start year (1995) to the end year (2005) to get:

• Given function: \( R = R_0 e^{0.15t} \)
• Integration over time gives: \[\int_0^{10} R_0 e^{0.15t} \, dt = R_0 \left[ \frac{e^{0.15t}}{0.15} \right]_0^{10}\]

By solving this integral, we find out how the function accumulates over the ten-year period. The solution to this integration gives us the total number of widgets sold during the specified time frame.
This method is crucial because it accounts for continuous growth, providing a more accurate representation of the total sales as opposed to just adding up annual figures.

Rate of Change

The rate of change is a critical concept when dealing with continuous growth models like exponential functions.
It describes how quickly a quantity (such as sales) is changing over time.

• In an exponential function, the rate of change is proportional to the current amount,
• Meaning as more widgets are sold, the rate at which they are sold increases.

In the exercise, this is represented by the function \( R = R_0 e^{0.15t} \).
The constant \( 0.15 \) indicates the rate at which the change occurs annually — a 15% increase each year, showcasing how rapidly sales numbers can grow. This feature makes exponential functions incredibly powerful for modeling processes that evolve over time with a consistent, percentage-based acceleration, making them suitable for tracking sales or population growth.

Initial Value Problem

An initial value problem helps us to understand the beginnings of changes over time.
Typically represented by differential equations, these problems provide a snapshot of the start condition and use it to predict future outcomes.
In this exercise, two different initial rates are considered for widgets sold per year (1000 and 150 million). Each sets up a different initial value problem:

• For 1000 widgets per year: Solving the initial value problem gives us insight into the pattern of sales if the starting scenario is modest.
• For 150 million widgets per year: This provides a trajectory of sales for a large-scale operation.

Solving these initial value problems involves integrating the given rate function with respective initial conditions to determine the sales over time.
By defining an initial condition, we can accurately model the exponential growth and assess different sales strategies or predict future outcomes based on initial conditions.