Problem 47 - Applied Calculus

Question

The cost of maintaining a home generally increases as the home becomes older. Suppose that the maintenance costs increase at the rate of \(1800 e^{0.05 x}\) (dollars per year) when the home is \(x\) years old. a. Find a formula for the total maintenance cost during the first \(x\) years. (Total maintenance should be zero at \(x=0 .)\) b. Use your answer to part (a) to find the total maintenance cost during the first 5 years.

Step-by-Step Solution

Answer

The total maintenance cost over 5 years is approximately $13,025.

Step 1: Understand the provided rate function

The exercise provides the rate of maintenance cost increase as a function of the age of the house. This rate is given by \( f(x) = 1800 e^{0.05x} \). This means the maintenance cost per year at year \( x \) is \( 1800 e^{0.05x} \).

Step 2: Set up the integral for total maintenance cost

To find the total maintenance cost over the first \( x \) years, we need to integrate the rate function from \( 0 \) to \( x \). The total maintenance cost function, \( C(x) \), is given by the integral \[ C(x) = \int_{0}^{x} 1800 e^{0.05t} \, dt. \]

Step 3: Integrate the rate function

To solve \( \int 1800 e^{0.05t} \, dt \), use the substitution \( u = 0.05t \), which gives \( du = 0.05dt \) or \( dt = 20 \, du \). Substituting, the integral becomes \[ \int 1800 e^{u} \cdot 20 \, du = 36000 \int e^{u} \, du = 36000 e^{u} + C. \] Substituting back \( u = 0.05t \), we have \[ 36000 e^{0.05t} + C. \]

Step 4: Evaluate the integral limits

Apply the limits from 0 to \( x \): \[ C(x) = \left[36000 e^{0.05t}\right]_0^x = 36000 e^{0.05x} - 36000 e^0. \] Simplify it to get \[ C(x) = 36000 e^{0.05x} - 36000. \]

Step 5: Calculate total cost over 5 years

Substitute \( x = 5 \) into the formula \( C(x) = 36000 e^{0.05x} - 36000 \). We get \( C(5) = 36000 e^{0.25} - 36000 \). Calculate the expression: \( C(5) \approx 36000 \times 1.284025 - 36000 \approx 13025. \)

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They are commonly seen in scenarios involving growth or decay processes, like population growth, radioactive decay, or, as in our example, cost increases over time.
In this exercise, our base is the natural constant Euler's number, denoted as \( e \), and the exponent is a function of the variable \( x \), specifically \( 0.05x \). This means the maintenance costs increase exponentially as the house ages.
The role of \( e \) is crucial because it models continuous growth or decay processes perfectly. The function given, \( 1800 e^{0.05x} \), tells us that not only does the cost increase, but it does so at a rate that grows faster as time progresses. Understanding this can help in predicting long-term costs and setting aside budgeting early on.

Integration Techniques

Integration is a core concept in calculus used to find the total value accumulated by adding up small incremental changes. In our example, the small increments are the increasing maintenance costs over time. By integrating the provided rate of cost increase, we can find the total maintenance cost during a certain period.
When dealing with exponential functions like \( 1800 e^{0.05x} \), a common technique is substitution. We set \( u = 0.05t \), simplifying the integral and allowing it to be easily evaluated. This method transforms a complex-looking integral into a basic form that is straightforward to solve.
After solving, the resulting integral gives us the function \( C(x) \), representing the total costs over a period from initial time to \( x \) years. It's derived by applying limits to the indefinite integral, ensuring we account for the exact timeframe desired.

Applications of Calculus

Calculus is not just a theoretical tool but has numerous practical applications. In real estate and home maintenance, calculus helps in predicting costs and analyzing trends over time. This exercise shows how you can apply integration to solve real-world problems, such as forecasting the financial burden of house repairs.
By integrating the cost rate function, we convert instantaneous costs into a comprehensive total, from which homeowners can make informed financial decisions. This approach also highlights one of calculus's strengths: translating changing scenarios into predictable outcomes using mathematical models.
Such applications are pivotal. They guide decisions in budgeting, investments, and even negotiating insurance terms. Knowing the total expected cost helps in planning and potentially saving funds for future needs, making calculus a powerful ally in strategic planning.

Cost Analysis

Cost analysis in this context focuses on understanding and forecasting the total expected maintenance expenses of a house over time. It involves examining the rate at which costs accrue, which here is initially given in an exponential form.
To perform cost analysis effectively, the first step is to model the cost rate as a mathematical function. In our case, it’s \( 1800 e^{0.05x} \). By integrating, as done in the solution, we determine the cumulative cost over desired periods, like the first five years.
This not only allows current homeowners to foresee financial obligations but also assists potential buyers in evaluating potential investment risks. Calculating the total maintenance over time can reveal how sustainable a home investment is and what kind of return one might expect, effectively guiding buying or selling decisions.