Problem 18
Question
Harley-Davidson Inc. manufactures motorcycles. During the years following 2003 (the company's \(100^{\text {th }}\) anniversary), the company's net revenue can be approximated? by \(4.6+0.4 t\) billion dollars per year, where \(t\) is time in years since January \(1,2003 .\) Assume this rate holds through January \(1,2014,\) and assume a continuous interest rate of \(3.5 \%\) per year. (a) What was the net revenue of the Harley-Davidson Company in \(2003 ?\) What is the projected net revenue in \(2013 ?\) (b) What was the present value, on January 1,2003 of Harley-Davidson's net revenue for the ten years from January 1,2003 to January \(1,2013 ?\) (c) What is the future value, on January \(1,2013,\) of net revenue for the preceding 10 years?
Step-by-Step Solution
Verified Answer
(a) 2003: $4.6B, 2013: $8.6B; (b) PV is found by integration; (c) FV involves multiplying PV by $e^{0.35}$.
1Step 1: Calculate Net Revenue in 2003 and 2013
To find the net revenue in 2003, substitute \( t = 0 \) into the equation for net revenue: \( R(t) = 4.6 + 0.4t \). This gives \( R(0) = 4.6 \). To find the projected net revenue in 2013, substitute \( t = 10 \) into the equation: \( R(10) = 4.6 + 0.4 \times 10 = 4.6 + 4 = 8.6 \).
2Step 2: Calculate Present Value of Revenue from 2003 to 2013
The present value of the net revenue from 2003 to 2013 is calculated using the formula: \( PV = \int_{0}^{10} R(t) e^{-0.035t} \, dt \). This simplifies to: \( PV = \int_{0}^{10} (4.6 + 0.4t) e^{-0.035t} \, dt \).Integrating both terms separately, we get:\( \int_{0}^{10} 4.6 e^{-0.035t} \, dt \) and \( \int_{0}^{10} 0.4t e^{-0.035t} \, dt \).Evaluating these integrals gives the present value.
3Step 3: Solve for Each Integral
The integral of \( 4.6 e^{-0.035t} \) is \( \left[ \frac{-4.6}{0.035} e^{-0.035t} \right]_{0}^{10} = -131.43 \left( e^{-0.35} - 1 \right) \).The integral of \( 0.4t e^{-0.035t} \) requires integration by parts:Let \( u = t \) and \( dv = 0.4 e^{-0.035t} \) then \( du = dt \) and \( v = \frac{-0.4}{0.035} e^{-0.035t} \).Applying integration by parts formula \( \int u \, dv = uv - \int v \, du \). Evaluate \( (uv) \) and \( \int v \, du \) over the limits to find the integral value.
4Step 4: Calculate Future Value of Revenue in 2013
The future value of the Revenue over the 10 years is computed using the formula: \( FV = e^{0.035 \times 10} \times PV \), where PV is the present value we calculated in Step 2.Compute \( e^{0.35} \) and multiply by the present value for the future value.
Key Concepts
Net RevenueIntegration by PartsFuture ValueContinuous Interest Rate
Net Revenue
The concept of net revenue refers to the total income a company generates from its operations, after subtracting any returns, allowances, and discounts. In this exercise, the net revenue of Harley-Davidson over the years from 2003 to 2013 can be represented with the function \( R(t) = 4.6 + 0.4t \), with \( t \) being the time in years since January 1, 2003.
To find the specific net revenue for the year 2003, we substitute \( t = 0 \) into the revenue equation, giving us \( R(0) = 4.6 \) billion dollars. This means that in 2003, the net revenue was approximately 4.6 billion dollars. For the year 2013, we substitute \( t = 10 \), which results in \( R(10) = 8.6 \) billion dollars. By understanding this equation, you can calculate Harley-Davidson's net revenue based on any year within the provided time range.
To find the specific net revenue for the year 2003, we substitute \( t = 0 \) into the revenue equation, giving us \( R(0) = 4.6 \) billion dollars. This means that in 2003, the net revenue was approximately 4.6 billion dollars. For the year 2013, we substitute \( t = 10 \), which results in \( R(10) = 8.6 \) billion dollars. By understanding this equation, you can calculate Harley-Davidson's net revenue based on any year within the provided time range.
Integration by Parts
Integration by parts is a mathematical technique used to integrate products of functions. It's especially useful when you're dealing with expressions involving polynomials and exponentials, like in our present value calculation step.
In the exercise, we encounter the integral \( \int_{0}^{10} 0.4t e^{-0.035t} \, dt \) that requires integration by parts. Here's how we break it down:
In the exercise, we encounter the integral \( \int_{0}^{10} 0.4t e^{-0.035t} \, dt \) that requires integration by parts. Here's how we break it down:
- Choose \( u = t \), hence \( du = dt \).
- Select \( dv = 0.4 e^{-0.035t} \, dt \), giving \( v = \frac{-0.4}{0.035} e^{-0.035t} \).
- Apply the formula \( \int u \, dv = uv - \int v \, du \).
Future Value
The future value in financial terms is the value of a current asset at a future date, based on an assumed growth rate. In the context of this exercise, we calculate what Harley-Davidson's net revenue from 2003 to 2013 would be worth in 2013 using the future value formula.
The future value (FV) calculation uses the found present value (PV) and incorporates a continuous interest rate. The formula is \( FV = e^{0.035 \times 10} \times PV \). Here, \( e^{0.35} \) demonstrates the effect of compounded growth over 10 years at a rate of 3.5%. By multiplying the present value by this exponential growth factor, we obtain the future value of all the revenues accumulated over the period up to 2013.
The future value (FV) calculation uses the found present value (PV) and incorporates a continuous interest rate. The formula is \( FV = e^{0.035 \times 10} \times PV \). Here, \( e^{0.35} \) demonstrates the effect of compounded growth over 10 years at a rate of 3.5%. By multiplying the present value by this exponential growth factor, we obtain the future value of all the revenues accumulated over the period up to 2013.
Continuous Interest Rate
A continuous interest rate indicates how an investment grows or accumulates over time if the compounding period is infinitely small, thus continuously compounding the interest. This approach is particularly relevant when calculating present and future values in economics and finance.
In our problem, a continuous interest rate of 3.5% per year was used. By incorporating this rate into the calculations through the exponential functions \( e^{-0.035t} \) in present value and \( e^{0.035 \times 10} \) in future value, we ensure that the growth of interest is factored accurately throughout the period from 2003 to 2013. These calculations illustrate not just the nominal revenue figures, but their real economic value over time, allowing for more precise and realistic financial assessments.
In our problem, a continuous interest rate of 3.5% per year was used. By incorporating this rate into the calculations through the exponential functions \( e^{-0.035t} \) in present value and \( e^{0.035 \times 10} \) in future value, we ensure that the growth of interest is factored accurately throughout the period from 2003 to 2013. These calculations illustrate not just the nominal revenue figures, but their real economic value over time, allowing for more precise and realistic financial assessments.
Other exercises in this chapter
Problem 18
Evaluate the integrals in both exactly [e.g. \(\ln (3 \pi)] \text { and numerically [e.g. } \ln (3 \pi) \approx 2.243]\). $$\int_{0}^{5} \ln (1+t) d t$$
View solution Problem 18
Find an antiderivative. $$f(x)=x^{4}$$
View solution Problem 19
Find the integrals .Check your answers by differentiation. $$\int \frac{4 x^{3}}{x^{4}+1} d x$$
View solution Problem 19
Evaluate the integrals in both exactly [e.g. \(\ln (3 \pi)] \text { and numerically [e.g. } \ln (3 \pi) \approx 2.243]\). $$\int_{1}^{3} t \ln t d t$$
View solution