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Chapter 2: Problem 18

The manager of an employee health plan for a firm has studied the balance \(B\), in millions of dollars, in the plan account as a function of \(t\), the number of years since the plan was instituted. He has determined that the account balance is given by the formula \(B=60+7 t-50 e^{0.1 t}\). a. Make a graph of \(B\) versus \(t\) over the first 7 years of the plan. b. At what time is the account balance at its maximum? c. What is the smallest value of the account balance over the first 7 years of the plan?

Short Answer

Expert verified
The account balance reaches its maximum at \( t \approx 3.059 \), with a minimum balance of \( B(7) \approx 53.87 \) million dollars.

Step by step solution

01

Understand the Function

The account balance is given by the function \( B(t) = 60 + 7t - 50e^{0.1t} \). Your task is to analyze this function based on the problem's requirements for the first 7 years.
02

Plot the Graph of B versus t

To make the graph, calculate the values of \( B(t) \) for each year from \( t = 0 \) to \( t = 7 \). Use these points to plot the graph. This step helps visualize how the account balance changes over time.
03

Find the Derivative

Find \( \frac{dB}{dt} \) to determine when \( B(t) \) is at its maximum or minimum. The derivative is: \[ \frac{dB}{dt} = 7 - 5e^{0.1t} \]
04

Solve for Critical Points

Set the derivative equal to zero to find critical points: \[ 7 - 5e^{0.1t} = 0 \] Solve for \( t \) to find \( t = \ln\left( \frac{7}{5} \right) / 0.1 \). This gives \( t \approx 3.059 \).
05

Determine Maximum Value

Evaluate \( B(t) \) at \( t = 3.059 \) and at the endpoints \( t = 0 \) and \( t = 7 \) to find the maximum balance. - \( B(0) = 60 - 50 = 10 \) - \( B(3.059) \approx 60 + 3.059 \times 7 - 50e^{0.1 \times 3.059} \approx 67.573 \) - \( B(7) = 60 + 49 - 50e^{0.7} \approx 53.87 \) The maximum value is \( B(3.059) \approx 67.573 \).
06

Determine Minimum Value

Compare the values calculated in Step 5 to determine the minimum balance over the 7 years. The smallest value is \( B(7) \approx 53.87 \), found by comparing \( B(0), B(3.059), \) and \( B(7) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Analysis
Function analysis is essential for understanding any mathematical function. Here, we're looking at the account balance function, given by: \[ B(t) = 60 + 7t - 50e^{0.1t} \] This function is a combination of linear and exponential expressions. The part \( 60 + 7t \) suggests a line increasing with time, while \( -50e^{0.1t} \) introduces a curve that decreases exponentially. Analyzing a function involves breaking it down to understand its behavior, growth rate, and long-term trends. For instance, at \( t = 0 \), \( B(t) = 10 \), starting with a decrease in initial conditions due to the negative exponential component. Over time, the linear growth dominates until the exponential deceleration is less significant. This interplay between linear growth and exponential decay is key in understanding how the balance in the account changes over time.
Derivatives
Derivatives help us understand how a function changes and are crucial for finding maximum or minimum values. Here, we calculate the first derivative of the balance function to analyze its rate of change. The first derivative of our function \( B(t) = 60 + 7t - 50e^{0.1t} \) is: \[ \frac{dB}{dt} = 7 - 5e^{0.1t} \]This derivative tells us how the account balance changes with respect to time. The term \( 7 \) represents a constant rate of increase, while \( -5e^{0.1t} \) is an exponential term that slows down the growth. To find when the balance is increasing or decreasing, we set this derivative to zero to find the critical points. This process helps identify turning points, where the function switches from increasing to decreasing, or vice versa.
Critical Points
Critical points in a function are where the derivative is zero or undefined, indicating potential maximum or minimum values. For our function, the critical point occurs when:\[ 7 - 5e^{0.1t} = 0 \]Solving this, we find:\[ t = \frac{\ln\left( \frac{7}{5} \right)}{0.1} \approx 3.059 \]At this critical point \( t = 3.059 \), the balance might be at a maximum or minimum. To verify, we compare this point with the values at the endpoints \( t = 0 \) and \( t = 7 \). Evaluating \( B(t) \) at these points reveals that \( t = 3.059 \) corresponds to the maximum account balance. Critical points often dictate the behavior of functions at key intervals, helping identify places of interest in a graph, such as peaks or troughs.
Graphing Functions
Graphing functions visually represents how values change over time. For the function \( B(t) = 60 + 7t - 50e^{0.1t} \), graphing requires calculating several values between \( t = 0 \) and \( t = 7 \). Here鈥檚 how you can approach this:
  • Start with endpoints: \( B(0) = 10 \) and \( B(7) \approx 53.87 \).
  • Include the critical point: \( B(3.059) \approx 67.573 \).
  • Plot these points, and sketch a smooth curve connecting them.
The curve peaks around \( t = 3.059 \), indicating the maximum balance, then smoothly declines towards \( t = 7 \). Graphs provide a clear picture of the function's behavior over time, revealing trends that aren't readily apparent from calculations alone. These visual tools are invaluable in finance to quickly understand the dynamics of accounts and investments.

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Most popular questions from this chapter

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