Problem 33 - Applied Calculus

Question

During a recession a firm's revenue declines continuously so that the revenue, \(R\) (measured in millions of dollars), in \(t\) years' time is given by \(R=5 e^{-0.15 t}\). (a) Calculate the current revenue and the revenue in two years' time. (b) After how many years will the revenue decline to \(\$ 2.7\) million?

Step-by-Step Solution

Answer

(a) Current revenue is 5 million, revenue in two years is 3.704 million. (b) Revenue declines to 2.7 million in approximately 4.108 years.

Step 1: Calculate the Current Revenue

To find the current revenue, we need to determine the revenue at time zero, i.e., when \( t = 0 \). Substitute \( t = 0 \) into the revenue formula: \[ R = 5 e^{-0.15 \times 0} = 5 \times e^0 = 5 \times 1 = 5. \]Therefore, the current revenue is 5 million dollars.

Step 2: Calculate the Revenue in Two Years

To find the revenue in two years, substitute \( t = 2 \) into the revenue formula:\[ R = 5 e^{-0.15 \times 2} = 5 e^{-0.3}. \]Use a calculator to find \( e^{-0.3} \approx 0.7408 \). Thus, \[ R = 5 \times 0.7408 \approx 3.704. \]Therefore, the revenue in two years' time is approximately 3.704 million dollars.

Step 3: Solve for the Decline in Revenue to $2.7 Million

We need to find \( t \) when the revenue \( R \) is 2.7 million dollars. Substitute \( R = 2.7 \) into the revenue equation:\[ 2.7 = 5 e^{-0.15 t}. \]Divide both sides by 5:\[ e^{-0.15 t} = \frac{2.7}{5}. \]Calculate \( \frac{2.7}{5} \approx 0.54 \). Now solve for \( t \) by taking the natural logarithm of both sides:\[ \ln(e^{-0.15 t}) = \ln(0.54). \]Using the property \( \ln(e^x) = x \), we get:\[ -0.15 t = \ln(0.54). \]Calculate \( \ln(0.54) \approx -0.6162 \). Thus,\[ -0.15 t = -0.6162. \]Divide both sides by -0.15:\[ t \approx \frac{-0.6162}{-0.15} \approx 4.108. \]Therefore, the revenue will decline to 2.7 million dollars in approximately 4.108 years.

Key Concepts

Continuous Revenue Decline

In the context of economics, continuous revenue decline refers to a situation where a company's revenue decreases steadily over time. This is often modeled with an exponential decay function, which helps predict future profits based on current data. During a recession, firms often experience such declines. The revenue formula provided, \(R = 5e^{-0.15t}\), is an example of how economists represent this decline mathematically.

• \(R\) represents revenue measured in millions of dollars.
• The exponential term \(e^{-0.15t}\) captures how revenue decreases at a continuous rate.
• Exponential decay equations like this help businesses forecast financial paths and strategize accordingly.

Understanding these models empowers businesses to make informed decisions during difficult economic times. The formula's components reveal key insights about the pace and pattern of financial decline.

Natural Logarithm

Natural logarithms, denoted as \(\ln\), are logarithms with base \(e\), where \(e\) is approximately 2.71828. They are crucial in solving equations involving exponential functions, especially when working with decay or growth models in economics.Using natural logarithms can simplify the process of isolating variables in exponential equations. For example, in solving for the time \(t\) when revenue declines to a specific amount, it's essential to take natural logarithms:

• The formula \( \ln(e^{-0.15t}) = \ln(0.54) \) helps isolate \( t \).
• Properties of logarithms, such as \( \ln(e^x) = x \), allow transformations that solve for unknowns easily.

Natural logarithms are powerful tools in both mathematics and economics for making complex equations more manageable. They allow you to express exponential decay models, like our revenue problem, in terms of linear relationships.

Revenue Calculation

Calculating revenue is a fundamental task in assessing a business’s financial health, particularly during unstable economic periods. Using mathematical models, such as the given function, helps one predict future earnings accurately.For current and future projections:

• The initial revenue at \(t = 0\) is straightforward, found by substituting zero: \(R = 5e^0 = 5\).
• For future revenue, replace \(t\) with the specific year: \(R = 5e^{-0.3}\) gives future values using exponential decay.

These calculations reveal essential patterns in revenue streams, allowing companies to adjust their financial strategies responsibly. Understanding these computations means interpreting real-world data through mathematical expressions.

Exponential Function in Economics

Exponential functions describe situations where quantities increase or decrease rapidly. In economics, these functions are crucial for modeling behaviors such as growth or decline, especially regarding revenue and expenses.The function \(R = 5e^{-0.15t}\) is typical of how exponential decay models are applied in economic scenarios. Key uses in economics include:

• Forecasting long-term revenue or cost trends.
• Assessing the impacts of economic changes, like a recession.

Exponential functions are powerful because they represent real-world unpredictability and change with mathematical precision. In our exercise, the given model helps determine not only current revenue but also insights into how quickly revenues will decline over time. Understanding how exponential functions operate allows businesses to navigate economic challenges more effectively.