Problem 66
Question
Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ("depreciates") through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A newspaper buys a printing press for \(\$ 800,000\) and estimates its useful life to be 20 years, after which its scrap value will be \(\$ 60,000 .\) Use the formula above Exercise 65 to find a formula for the value \(V\) of the press after \(t\) years, for \(0 \leq t \leq 20\) b. Use your formula to find the value of the press after 10 years. c. Graph the function found in part (a) on a graphing calculator on the window [0,20] by [0,800,000] . [Hint: Use \(x\) instead of \(t .]\)
Step-by-Step Solution
Verified Answer
The press's value after 10 years is $430,000.
1Step 1: Understand the Base Formula
The straight-line depreciation formula is used to determine the asset's value after a given number of years. It is given as: \[ V = P - \left(\frac{P - S}{L}\right) \cdot t \] where \( V \) is the value of the asset after \( t \) years, \( P \) is the original price, \( S \) is the scrap value, and \( L \) is the useful lifetime.
2Step 2: Identify Given Values
For the newspaper printing press:- Original Price \( (P) = \\(800,000 \)- Scrap Value \( (S) = \\)60,000 \)- Useful Lifetime \( (L) = 20 \text{ years} \)
3Step 3: Substitute Given Values into the Formula
Substitute the identified values into the depreciation formula:\[ V = 800,000 - \left(\frac{800,000 - 60,000}{20}\right) \cdot t \]
4Step 4: Simplify the Equation
First calculate the fraction part: - \( 800,000 - 60,000 = 740,000 \) - \( \frac{740,000}{20} = 37,000 \)Thus, the formula simplifies to:\[ V = 800,000 - 37,000t \]
5Step 5: Calculate Value after 10 Years
Substitute \( t = 10 \) into the equation:\[ V = 800,000 - 37,000 \cdot 10 \]\[ V = 800,000 - 370,000 \]\[ V = 430,000 \]So, the value of the press after 10 years is \( \$430,000 \).
6Step 6: Graph the Function
Using a graphing calculator, enter the function \( V = 800,000 - 37,000x \) where \( x \) substitutes for \( t \). Set the window to [0,20] for the \( x \)-axis and [0,800,000] for the \( y \)-axis to visualize the depreciation over the press's 20-year life.
Key Concepts
Asset ValuationDepreciation FormulaGraphing CalculatorsCalculus in EconomicsMathematical Modeling
Asset Valuation
Asset valuation is an important concept when it comes to understanding how the value of a company's assets, like equipment or machinery, changes over time. It's crucial for businesses to know the value of their assets for financial reporting and decision-making.
Asset valuation allows companies to determine how much their assets are worth at any point in their useful lifetime. It helps in planning and evaluating business activities. When businesses understand the depreciation of their assets, they can make better decisions regarding when to replace equipment or invest in maintenance.
In essence, asset valuation provides a snapshot of the financial health of a business, enabling stakeholders to make informed economic decisions. This is particularly important when assets, like the newspaper's printing press in our example, have a predetermined useful life and scrap value.
Asset valuation allows companies to determine how much their assets are worth at any point in their useful lifetime. It helps in planning and evaluating business activities. When businesses understand the depreciation of their assets, they can make better decisions regarding when to replace equipment or invest in maintenance.
In essence, asset valuation provides a snapshot of the financial health of a business, enabling stakeholders to make informed economic decisions. This is particularly important when assets, like the newspaper's printing press in our example, have a predetermined useful life and scrap value.
Depreciation Formula
The depreciation formula we are using here is specifically designed to calculate straight-line depreciation. This method reduces the asset's value equally across its useful life. The formula is:
- \[ V = P - \left(\frac{P - S}{L}\right) \cdot t \]
- \(V\) represents the value of the asset at time \(t\).
- \(P\) is the original purchase price of the asset.
- \(S\) is the scrap value of the asset after its useful lifetime.
- \(L\) is the useful lifetime of the asset.
- \(t\) is the number of years since the asset was purchased.
Graphing Calculators
Graphing calculators are powerful tools that help visualize mathematical relationships. In the case of the straight-line depreciation, it translates abstract formulas into a clear visual representation.
When we input the formula \( V = 800,000 - 37,000x \) into the calculator, where \( x \) represents years, the graph shows a straight line. This line starts at the original value of the asset and gradually decreases each year until it reaches the scrap value at the end of its 20-year life.
Using graphing calculators, students can set specific windows, like \([0, 20]\) on the \(x\)-axis and \([0, 800,000]\) on the \(y\)-axis, to better understand the depreciation trend over time, enabling a practical understanding of how the asset loses value.
When we input the formula \( V = 800,000 - 37,000x \) into the calculator, where \( x \) represents years, the graph shows a straight line. This line starts at the original value of the asset and gradually decreases each year until it reaches the scrap value at the end of its 20-year life.
Using graphing calculators, students can set specific windows, like \([0, 20]\) on the \(x\)-axis and \([0, 800,000]\) on the \(y\)-axis, to better understand the depreciation trend over time, enabling a practical understanding of how the asset loses value.
Calculus in Economics
Calculus plays an important role in economics, offering insights into how things change and move over time, which is particularly useful in the context of asset depreciation.
While straight-line depreciation is a linear model, calculus can be applied to more complex depreciation methods, like declining balance or sum-of-the-years-digits, where the rate of depreciation changes over time. Calculus allows economists to analyze these changing rates and predict future asset values more accurately.
Understanding the basic principles of calculus helps economic students and professionals alike describe and solve real-world financial problems, like depreciation, with greater precision, increasing the effectiveness of economic decision-making.
While straight-line depreciation is a linear model, calculus can be applied to more complex depreciation methods, like declining balance or sum-of-the-years-digits, where the rate of depreciation changes over time. Calculus allows economists to analyze these changing rates and predict future asset values more accurately.
Understanding the basic principles of calculus helps economic students and professionals alike describe and solve real-world financial problems, like depreciation, with greater precision, increasing the effectiveness of economic decision-making.
Mathematical Modeling
Mathematical modeling is a technique used to represent real-world situations through mathematical formulas and concepts, such as the straight-line depreciation formula used in our exercise.
This approach allows practitioners to simulate and predict how assets will depreciate over time, under certain conditions. In the case of straight-line depreciation, the model assumes that the asset will lose the same amount of value each year. Though it simplifies reality, the model remains immensely useful for accounting and planning purposes.
By understanding mathematical modeling, students and professionals can gain insights into how theoretical models represent and impact the economics of everyday life, facilitating better planning and financial forecasting.
This approach allows practitioners to simulate and predict how assets will depreciate over time, under certain conditions. In the case of straight-line depreciation, the model assumes that the asset will lose the same amount of value each year. Though it simplifies reality, the model remains immensely useful for accounting and planning purposes.
By understanding mathematical modeling, students and professionals can gain insights into how theoretical models represent and impact the economics of everyday life, facilitating better planning and financial forecasting.
Other exercises in this chapter
Problem 66
Write each expression in power form \(a x^{b}\) for numbers \(a\) and \(b\). $$ \frac{3 \sqrt{x}}{x} $$
View solution Problem 66
BUSINESS: Cost Functions A company manufactures bicycles at a cost of \(\$ 55\) each. If the company's fixed costs are \(\$ 900\), express the company's costs a
View solution Problem 67
Write each expression in power form \(a x^{b}\) for numbers \(a\) and \(b\). $$ \frac{12 \sqrt[3]{x^{2}}}{3 x^{2}} $$
View solution Problem 67
BUSINESS: Salary An employee's weekly salary is \(\$ 500\) plus \(\$ 15\) per hour of overtime. Find a function \(P(x)\) giving his pay for a week in which he w
View solution