Problem 63 - Precalculus: Mathematical for Calculus

Question

A construction company purchases a bulldozer for 160,000 dollars. Each year the value of the bulldozer depreciates by \(20 \%\) of its value in the preceding year. Let \(V_{n}\) be the value of the bulldozer in the \(n\) th year. (Let \(n=1\) be the year the bulldozer is purchased.) (a) Find a formula for \(V_{n}\) (b) In what year will the value of the bulldozer be less than 100,000 dollars?

Step-by-Step Solution

Answer

(a) \(V_n = 160,000 \times (0.8)^n\); (b) 3rd year

Step 1: Understanding Depreciation

The bulldozer's value is reduced by \(20\%\) each year. This means by the end of each year, the value is \(80\%\) of its value at the beginning of the year.

Step 2: Establishing a Formula for Depreciation

Since the value decreases by \(20\%\), the value at the end of a year is \(80\%\) of the starting value. Mathematically, this is modeled by multiplying by \(0.8\). If the initial value is \(V_0 = 160,000\), the value at year \(n\) can be expressed as: \[ V_n = 160,000 \times (0.8)^n \]

Step 3: Calculate Value for Specific Years to Find When It's Less Than $100,000

We need to determine the smallest \(n\) such that \(V_n < 100,000\). Using our formula from Step 2:\[ 160,000 \times (0.8)^n < 100,000 \] Divide both sides by \(160,000\) to get: \[ (0.8)^n < \frac{100,000}{160,000} \] \[ (0.8)^n < 0.625 \]

Step 4: Solve the Inequality for n

To solve \((0.8)^n < 0.625\), take the logarithm of both sides:\[ n \cdot \log(0.8) < \log(0.625) \] Solving for \(n\), we have:\[ n > \frac{\log(0.625)}{\log(0.8)} \] Calculate the values:\(\log(0.625) \approx -0.2041\) and \(\log(0.8) \approx -0.09691\),so,\[ n > \frac{-0.2041}{-0.09691} \approx 2.11 \] Since \(n\) must be an integer, the smallest \(n\) is 3.

Key Concepts

Depreciation

Depreciation is a concept widely used in both accounting and financial mathematics. It's crucial to understand that depreciation refers to the reduction in the value of an asset over time. This decrease in value can be due to various factors like wear and tear, obsolescence, or age. For tangible assets such as vehicles or machinery, such as the bulldozer from our example, depreciation is often predicted as a percentage.

• The bulldozer's value decreases by 20% each year.
• After each year, it retains 80% of its previous year's value.

In practical terms, this means multiplying the current value by 0.8 to determine its subsequent year's value. Mathematically, this is expressed by the formula: \[ V_n = 160,000 \times (0.8)^n \] where \( V_n \) is the asset's value in the \( n \text{th} \) year. Understanding this formula allows us to estimate future depreciation, plan for replacements, and manage financial portfolios effectively.

Financial Mathematics

Financial mathematics involves the application of mathematical methods to solve problems in finance. One major application is understanding how asset values change over time. In our problem, this involves considering how an asset depreciates.

• Depreciation impacts an asset's financial value yearly.
• Mathematical modeling helps predict this decline effectively.

By considering the bulldozer's depreciation, investors or managers can make informed decisions about maintaining, selling, or replacing assets. Financial mathematics formulas, such as the exponential decay formula used here, help forecast when an asset reaches a particular value threshold. This allows for strategic planning, ensuring funds are appropriately allocated without unexpected financial strain. The exponential nature of the formula: \[ V_n = 160,000 \times (0.8)^n \] shows a rapid decrease initially and then levels out over time, a common trait in real-world financial applications.

Logarithms

Logarithms are an essential mathematical tool, particularly in solving exponential equations. They can simplify the process of identifying the time it takes for an asset to reach a particular value.

• The bulldozer's depreciation involves exponential decay: \[ V_n = 160,000 \times (0.8)^n \]
• To find the year when its value is below \\(100,000, an inequality is used.

By taking the logarithm of both sides, we transform the problem from an exponential equation into a linear form, which is simpler to solve. The calculated inequality:\[ n > \frac{\log(0.625)}{\log(0.8)} \] reveals when the value will be less than \\)100,000. Plugging in the logarithm values, we find \( n \approx 2.11 \). Therefore, the bulldozer will be worth less than \$100,000 after 3 years. Using logarithms thus gives clear insight into timeframes for financial decision-making.