Problem 103 - Calculus Volume 1

Question

[T] A company purchases some computer equipment for \(\mathrm{S} 20,500\) , At the end of a 3 -year period, the value of the equipment has decreased linearly to \(\$ 12,300\) . a. Find a function \(y=V(t)\) that determines the value V of the equipment at the end of \(t\) years. b. Find and interpret the meaning of the \(x\) - and \(y-\) intercepts for this situation. c. What is the value of the equipment at the end of 5 years? d. When will the value of the equipment be \(\$ 3000 ?\)

Step-by-Step Solution

Answer

a) The function is \( V(t) = -2,733.33t + 20,500 \). b) x-intercept: \( t \approx 7.5 \), y-intercept: \( \$20,500 \). c) Value at 5 years: \( \$6,833.35 \). d) Value \( \$3,000 \) at \( t \approx 6.4 \) years.

Step 1: Understanding a Linear Depreciation Function

The value of the equipment decreases linearly over time. A linear function can be expressed as \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Here, \( t \) represents time in years, and \( V(t) \) represents the equipment value at time \( t \). We must find the initial value \( b \) and the rate of depreciation \( m \).

Step 2: Calculating the Slope (Rate of Depreciation)

The value decreases from \( \\(20,500 \) to \( \\)12,300 \) over 3 years. The slope is given by the formula: \( m = \frac{\Delta y}{\Delta t} = \frac{12,300 - 20,500}{3 - 0} = \frac{-8,200}{3} = -\$2,733.33 \).

Step 3: Setting Up the Function

The initial value \( b \) is identical to the purchase price, \( \$20,500 \). The linear function representing the equipment's value is \( V(t) = -2,733.33t + 20,500 \).

Step 4: Finding the y-intercept

The y-intercept represents the value of the equipment at \( t = 0 \). This is simply \( \$20,500 \), the purchase price.

Step 5: Finding the x-intercept

The x-intercept occurs when the equipment's value is \( \$0 \):\[ 0 = -2,733.33t + 20,500 \]\[ 2,733.33t = 20,500 \]\[ t \approx 7.5 \text{ years}\].This means the equipment will be worthless after about 7.5 years.

Step 6: Calculating Value After 5 Years

Substitute \( t = 5 \) into \( V(t) = -2,733.33t + 20,500 \):\[ V(5) = -2,733.33 \times 5 + 20,500 = -13,666.65 + 20,500 = \$6,833.35 \].

Step 7: Determining When the Equipment Value is \( \$3,000 \)

Set \( V(t) = 3,000 \) and solve for \( t \):\[ 3,000 = -2,733.33t + 20,500 \]\[ 2,733.33t = 20,500 - 3,000 \]\[ 2,733.33t = 17,500 \]\[ t \approx 6.4 \text{ years} \].The equipment will be worth \( \$3,000 \) after approximately 6.4 years.

Key Concepts

Linear Function

In mathematics, a linear function is often used to model relationships where there is a constant rate of change. It is expressed in the general form \( y = mx + b \), where \( m \) represents the slope of the line and \( b \) represents the y-intercept. For the problem of equipment depreciation, a linear function captures how the equipment value decreases steadily over time.

Here:

• \( y \) or \( V(t) \) is the function representing the equipment value at any time \( t \).
• \( t \) denotes time in years.
• \( m \), the slope, shows how quickly the value decreases.
• \( b \), the y-intercept, is the initial value – the purchase price.

This function helps determine future and past equipment value based on the time passed since the purchase.

Slope Calculation

To determine the rate at which the equipment's value decreases, you calculate the slope of the linear function. Slope illustrates the change in value per unit time. Using the formula \( m = \frac{\Delta y}{\Delta t} \), the slope gives a clear measure of how much the equipment value falls each year.

In our case, the slope \( m \) was calculated by:

• Subtracting the final value \( (12,300) \) from the initial value \( (20,500) \).
• Dividing by the change in time \( (3 \, \text{years}) \).

The result is a slope of \(-2,733.33\), meaning the equipment loses \( \$2,733.33 \) in value every year. Understanding the slope is essential for estimating how long it will take for the equipment's value to reach certain levels.

Intercepts in Mathematics

Intercepts are crucial points in understanding the behavior of linear functions in relation to real-world contexts.

**Y-Intercept:**
The y-intercept occurs where the line crosses the y-axis, meaning \( t = 0 \). It represents the initial value of the equipment when first purchased. Here, the y-intercept is \$20,500. This is the starting point of the value depreciation.

**X-Intercept:**
The x-intercept occurs where the line crosses the x-axis, or where \( y = 0 \). It signifies when the equipment's value will become zero. Solving for \( t \) when \( y = 0 \) gives the time it will take for the depreciation to completely eliminate the equipment's value. Here, the time is approximately after 7.5 years. Understanding intercepts allows you to effectively interpret key moments in the function's practical application.

Value of Equipment Over Time

Equipment value over time follows the course determined by its linear depreciation. This concept helps stakeholders make informed decisions about budgeting and resource allocation.

For instance, using the function \( V(t) = -2,733.33t + 20,500 \), we can:

• Find the equipment's current value at any given time, such as after 5 years, resulting in a value of \\(6,833.35.
• Estimate when it will reach specific values, like identifying it will reach \\)3,000 after about 6.4 years.
• Determine the equipment's lifespan, aligning financial strategies with depreciation rates.

By understanding and applying these concepts involving time and value, practical use and financial investment in equipment can be optimized efficiently.