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Capitalized cost Find the capitalized cost \(C\) of an asset (a) for \(n=5\) years, (b) for \(n=10\) years, and (c) forever. The capitalized cost is given by $$ C=C_{0}+\int_{0}^{n} c(t) e^{-r t} d t $$ where \(C_{0}\) is the original investment, \(t\) is the time in years, \(r\) is the annual interest rate compounded continuously, and \(c(t)\) is the annual cost of maintenance (measured in dollars). [Hint: For part (c), see Exercises \(35-38 .]\) $$ C_{0}=\$ 650,000, c(t)=25,000, r=10 \% $$

Short Answer

Expert verified
The capitalized cost for \( n = 5 \) years, \( n = 10 \) years, and \( n = \infty \) years, can be found by plugging the given values into the provided formula and correctly evaluating the integrals for each term. As the integral computation might be complex, a scientific calculator or computer algebra system may be helpful to get the final numerical values for the capitalized cost.

Step by step solution

01

Identify Given Variables

The variables that have been given in the problem statement are: The original investment \( C_0 = \$650,000 \), the annual cost of maintenance \( c(t) = \$25,000 \), and the annual interest rate is \( r = 10 \% \).
02

Calculate Capitalized Cost for 5 Years

We find the capitalized cost for \( n = 5 \) years as follows: Plug the provided values in the formula and evaluate, \( C = \$650,000 + \int_{0}^{5}\$25,000 e^{-0.10t} dt \). By solving this integral, we can find capitalized cost.
03

Calculate Capitalized Cost for 10 Years

For \( n = 10 \) years, the capitalized cost calculation follows a similar method: Again plug the values into the formula, \( C = \$650,000 + \int_{0}^{10}\$25,000 e^{-0.10t} dt \), and solve the integral to obtain the capitalized cost.
04

Calculate Capitalized Cost for Forever

For the asset being held indefinitely, i.e. \( n = \infty \), the capitalized cost becomes, \( C = \$650,000 + \int_{0}^{\infty}\$25,000 e^{-0.10t} dt \). As before, critically evaluating this integral will give the capitalized cost.

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Key Concepts

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

Continuous Compounding Interest
Continuous compounding interest is the most intense form of calculating interest, where interest is added to the principal continuously at every moment. Unlike simple or compound interest which is calculated periodically, in continuous compounding, the interest is calculated and added instantaneously. The formula for continuous compounding can be expressed using the natural exponential function:
\( A = Pe^{rt} \)
where \( P \) is the principal amount, \( r \) is the annual interest rate, \( t \) is the time in years, and \( e \) is the base of the natural logarithm (approximately equal to 2.71828). This concept is crucial for our capitalized cost problem because it defines how we discount the future costs of maintenance to their present value, accounting for the time value of money.
Cost of Maintenance Calculation
Calculating the cost of maintenance is another key component of determining the capitalized cost of an asset. The cost of maintenance, \( c(t) \), includes all necessary expenditures required to keep the asset functioning properly over time. In our problem, maintenance costs are constant at \( \$25,000 \) per year. However, to account for the time value of money due to continuous compounding interest, these costs are discounted back to their present value using the formula from the previous section. This is done by multiplying the maintenance costs by the discount factor \( e^{-rt} \) and integrating over the specified time period.
Integral Evaluation in Economics
The use of integrals in economics is prominent for problems involving continuous change over a period of time. Integral evaluation is the process of calculating the area under a curve, which, in economic terms, can represent the total cost, total revenue, or, as in our case, the capitalized cost. To find the present value of the continuous stream of maintenance costs, we need to evaluate the integral of the maintenance cost discounted by the continuous compounding factor over time.
\( \int_{0}^{n} c(t) e^{-rt} dt \)
By solving this integral, we factor in both the constant maintenance costs and their decreasing value over time due to the interest rate. This allows us to calculate accurately the present value of the asset’s future maintenance costs.
Present Value of Annuity
The present value of an annuity is the current worth of a series of future payments, discounted to account for the time value of money. In the context of our capitalized cost problem, the maintenance costs can be seen as an annuity—a series of fixed payments over the specified term. Calculating the present value of an annuity helps us understand how much those future maintenance payments are worth in today's dollars. It is the integration of these present values that form part of the total capitalized cost in our problem. As the time period approaches infinity, we essentially calculate the present value of a perpetuity, representing never-ending constant payments. This concept ties directly with our task to evaluate the integral for the capitalized cost over 5 years, 10 years, and perpetually.

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Most popular questions from this chapter

Capitalized cost Find the capitalized cost \(C\) of an asset (a) for \(n=5\) years, (b) for \(n=10\) years, and (c) forever. The capitalized cost is given by $$ C=C_{0}+\int_{0}^{n} c(t) e^{-r t} d t $$ where \(C_{0}\) is the original investment, \(t\) is the time in years, \(r\) is the annual interest rate compounded continuously, and \(c(t)\) is the annual cost of maintenance (measured in dollars). [Hint: For part (c), see Exercises \(35-38 .]\) $$ C_{0}=\$ 650,000, c(t)=25,000(1+0.08 t), r=12 \% $$

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