Problem 27 - Applied Calculus

Question

(a) Between 2000 and 2010 , ACME Widgets sold widgets at a continuous rate of \(R=R_{0} e^{0.125 t}\) widgets per year, where \(t\) is time in years since January 1 2000\. Suppose they were selling widgets at a rate of 1000 per year on January \(1,2000 .\) How many widgets did they sell between 2000 and \(2010 ?\) How many did they sell if the rate on January 1,2000 was 1,000,000 widgets per year? (b) In the first case ( 1000 widgets per year on January 1,2000)\(,\) how long did it take for half the widgets in the ten-year period to be sold? In the second case \((1,000,000 \text { widgets per year on January } 1,2000)\) when had half the widgets in the ten-year period been sold? (c) In \(2010,\) 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

They sold 19922 widgets with a 1000 rate and 19922000 with 1000000 rate. Half were sold after 10 years for both rates, thus widgets must last 10 years.

Step 1: Calculate the Integral for Total Widgets

Start by calculating the total number of widgets sold from 2000 to 2010. The rate function is given as \( R = R_0 e^{0.125 t} \). The total widgets sold will be the integral of \( R \) from \( t = 0 \) to \( t = 10 \). \[\text{Total} = \int_{0}^{10} R_0 e^{0.125 t} \, dt\]

Step 2: Solve the Integral

The integration of \( R_0 e^{0.125 t} \) with respect to \( t \) is:\[\frac{R_0}{0.125} e^{0.125 t} = 8R_0 e^{0.125 t}\]Calculate the total widgets sold as follows:\[8R_0 \left[ e^{0.125 \times 10} - e^{0.125 \times 0} \right] = 8R_0 \left[ e^{1.25} - 1 \right]\]

Step 3: Calculate Total Widgets Sold for R0 = 1000

Using \( R_0 = 1000 \):\[\text{Total} = 8 \times 1000 \times (e^{1.25} - 1) \approx 8 \times 1000 \times (3.4903 - 1) = 19922 \text{ widgets}\]

Step 4: Calculate Total Widgets Sold for R0 = 1000000

Using \( R_0 = 1000000 \):\[\text{Total} = 8 \times 1000000 \times (e^{1.25} - 1) \approx 8 \times 1000000 \times (3.4903 - 1) = 19922000 \text{ widgets}\]

Step 5: Find Time for Half Widgets Sold for R0 = 1000

To find when half the widgets are sold, solve:\[4000 (e^{0.125 t} - 1) = \frac{19922}{2}\]This simplifies to:\[e^{0.125 t} - 1 = \frac{9961}{4000}\]\[e^{0.125 t} = 3.49025\]Solve for \( t \):\[t = \frac{\ln(3.49025)}{0.125} \approx 10 \text{ years}\]

Step 6: Find Time for Half Widgets Sold for R0 = 1000000

Similar to the previous step, solve for \( t \) with \( R_0 = 1000000 \):\[4000000 (e^{0.125 t} - 1) = \frac{19922000}{2}\]This simplifies to:\[e^{0.125 t} = 3.49025\]\[t = \frac{\ln(3.49025)}{0.125} \approx 10 \text{ years}\]

Step 7: Evaluate Widget Longevity for 2010 Advertisement

Since in both scenarios half the widgets were sold at approximately the end of year 10, and since ACME claims half the widgets are still in use in 2010, each widget must last at least 10 years to justify the claim.

Key Concepts

Integration

When dealing with functions that change continuously over time, integration is a powerful tool to calculate the total or accumulated quantity over a given period. In the context of widget sales, the integration helps determine the total number of widgets sold over a decade.

To calculate the total widgets sold, we integrate the rate function, which is given by the equation: \[ R = R_0 e^{0.125 t} \]Integrating from year 0 to year 10 gives us: \[ \int_{0}^{10} R_0 e^{0.125 t} \, dt \]
This integral represents the sum of all widget sales over ten years. By solving this, we determine how many widgets were sold during this period, helping us understand the exponential increase in sales and how it accumulates over time.

Continuous Rate

The concept of continuous rate in this scenario refers to how ACME Widgets sold their products at a continuously increasing pace each year. This rate is modeled by the function:\[ R = R_0 e^{0.125 t} \]where \( R_0 \) is the initial rate of sales at the beginning of the period (January 1, 2000).

A continuous rate signifies that sales aren't happening in discrete steps but rather in a fluid, ongoing manner. Such a description reflects real-world scenarios more accurately, where sales and transactions occur constantly rather than just at specified intervals. This model allows us to predict the total sales over time by integrating the rate function, taking into account the continuous nature of sales growth.

Widget Sales

ACME's widget sales over the ten years from 2000 to 2010 are a prime example of exponential growth influenced by economic factors, improving technology, and market expansion. By using the given rate equation with different initial sales rates \( R_0 \), we calculate how many widgets were sold in total for specific scenarios.

For rate \( R_0 = 1000 \), the total widgets sold were approximately 19,922. When the initial rate was set to \( R_0 = 1,000,000 \), the number of widgets sold rose significantly to about 19,922,000. These calculations not only demonstrate exponential growth but also emphasize how initial conditions greatly impact total sales over time.

Understanding these sales dynamics helps businesses to plan, forecast future sales, and make informed decisions regarding production and marketing strategies.

Logarithmic Functions

Logarithmic functions are crucial for solving problems where the output exponentiates with time, such as finding out when a certain portion of widgets is sold. In this task, we use logarithms to determine when half of the total widgets were sold during the ten-year period.

For instance, given for the initial rate \( R_0 \), we need to know when sales reached half. For this, we solve:\[ e^{0.125 t} = 3.49025 \]Taking the natural logarithm on both sides helps to isolate \( t \):\[ t = \frac{\ln(3.49025)}{0.125} \approx 10 \text{ years} \]

Logarithmic functions make it feasible to go backward from an exponential model, allowing us to ascertain specific points in time such as when half of the widget inventory was sold. This process significantly aids in planning logistics and understanding sales life cycles.