Problem 71

Question

A construction company purchases a bulldozer for $\$ 160,000$. 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 $\$ 100,000 ?$

Step-by-Step Solution

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Answer

Formula: $ V_n = 160,000 \times 0.8^{n-1} $. Value < $100,000 in year 4.

1Step 1: Understand Depreciation

The value of the bulldozer decreases by 20% each year. This means after each year, the bulldozer retains 80% of its value from the previous year. Thus, the relationship can be expressed as a multiplication by a factor of 0.8.

2Step 2: Formulate the Recurrence Relation

To express the value of the bulldozer in the nth year, note that the initial value is $ V_1 = 160,000 $. For subsequent years, the value decreases to 80% of its previous year's value, so $ V_n = 0.8 \times V_{n-1} $.

3Step 3: Derive the Formula for V_n

Starting with $ V_1 = 160,000 $, the value in the nth year can be given by the formula \[ V_n = 160,000 \times 0.8^{n-1} \]. You derive this by repeatedly applying the depreciation reduction factor.

4Step 4: Find the Year When Value Falls Below $100,000

Set $ V_n < 100,000 $ to find when the value falls below this threshold. Substitute the formula to get \[ 160,000 \times 0.8^{n-1} < 100,000 \].

5Step 5: Solve the Inequality

Simplify the expression: \[ 0.8^{n-1} < \frac{100,000}{160,000} = 0.625 \]. To solve $ 0.8^{n-1} < 0.625 $, take the logarithm of both sides. For example, using base 10: \[ \log(0.8^{n-1}) < \log(0.625) \].

6Step 6: Calculate n

Using the log property $ \log(a^b) = b \cdot \log(a) $, rewrite as $ (n-1) \cdot \log(0.8) < \log(0.625) $. Then calculate: $ (n-1) > \frac{\log(0.625)}{\log(0.8)} $, giving $ n > 3.10 $. Hence, after 4 complete years, the value of the bulldozer will be below $100,000.

Key Concepts

Recurrence Relation in DepreciationUnderstanding Exponential DecaySolving the Inequality in Depreciation

Recurrence Relation in Depreciation

In the context of depreciation, understanding recurrence relations is crucial. A recurrence relation defines each term of a sequence as a function of its preceding terms. For example, the bulldozer's value after each year can be seen as a series of values, where each value depends on the previous year's value.

The recurrence relation for the bulldozer's depreciation is given by

$ V_n = 0.8 \times V_{n-1} $,
with the initial value, $ V_1 = 160,000 $.

This means every new year's value is 80% of its last year's value. The 0.8 comes from subtracting the 20% depreciation from 100%. By repeatedly applying this formula, we systematically compute subsequent values.

Understanding Exponential Decay

Exponential decay describes a process where quantities decrease over time at a rate proportional to their current value. In the case of the bulldozer, its value decreases exponentially each year due to the fixed percentage loss known as depreciation.

Let's see how this relates to the formula:

Initially, the bulldozer's value is $ 160,000 $.
Each year, the value multiplies by 0.8, similarly as seen in the recurrence relation.

This formula can be expressed as:

$ V_n = 160,000 \times 0.8^{n-1} $.

The $ 0.8^{n-1} $ represents the exponential part, showing that as $ n $ increases, the value decreases more due to continuous multiplication by 0.8.

Solving the Inequality in Depreciation

Solving inequalities in a depreciation context involves finding when the value of an asset falls below a particular threshold. For the bulldozer, we need to know when its value is less than $ 100,000 $.

Using the equation