Problem 38 - Intermediate Algebra

Question

Use linear functions. The linear depreciation method assumes that an item depreciates the same amount each year. Suppose a new piece of machinery costs \(\$ 32,500\) and it depreciates \(\$ 1950\) each year for \(t\) years. (a) Set up a linear function that yields the value of the machinery after \(t\) years. (b) Find the value of the machinery after 5 years. (c) Find the value of the machinery after 8 years. (d) Graph the function from part (a). (e) Use the graph from part (d) to approximate how many years it takes for the value of the machinery to become zero. (f) Use the function to determine how long it takes for the value of the machinery to become zero.

Step-by-Step Solution

Answer

(a) \(V(t) = 32500 - 1950t\); (b) \(\$22,750\); (c) \(\$16,900\); (e) About 17 years; (f) 16.67 years.

Step 1: Set Up the Linear Function

To create a linear function, recognize that the machine loses \(1950 in value each year. Begin with the initial value of \)32,500. The slope representing depreciation is -1950. Thus, the linear function is: \[ V(t) = 32500 - 1950t \]where \(V(t)\) is the value of the machinery after \(t\) years.

Step 2: Calculate Value After 5 Years

Substitute \(t = 5\) into the linear function:\[V(5) = 32500 - 1950 \times 5\]Calculate the value:\[V(5) = 32500 - 9750 = 22750\]The value of the machinery after 5 years is \( \$22,750 \).

Step 3: Calculate Value After 8 Years

Substitute \(t = 8\) into the linear function:\[V(8) = 32500 - 1950 \times 8\]Calculate the value:\[V(8) = 32500 - 15600 = 16900\]The value of the machinery after 8 years is \( \$16,900 \).

Step 4: Graph the Linear Function

Plot the function \(V(t) = 32500 - 1950t\) on a graph with \(t\) on the x-axis (years) and \(V(t)\) on the y-axis (value in dollars). Mark key points, such as \( t = 0 \) with value \( \\(32,500 \), and \( t = 5 \) with value \( \\)22,750 \), and connect to form a line.

Step 5: Estimate Years for Zero Value Using Graph

From the graph, observe where the line crosses the x-axis. This crossing point is where the value of the machinery becomes zero. It visually appears to be around 16 or 17 years.

Step 6: Calculate Exact Zero Value Year with Function

Set the linear function equal to zero and solve for \(t\):\[0 = 32500 - 1950t\]Re-arrange to solve for \(t\):\[1950t = 32500 \ t = \frac{32500}{1950} \ t \approx 16.67\]It will take approximately 16.67 years for the machinery's value to reduce to zero.

Key Concepts

Linear Depreciation

Linear depreciation is a straightforward method used to spread the cost of an asset uniformly over its useful life. In our exercise, the machinery initially costs \(32,500. \) Each year, the machinery loses \(1950, \) which is consistent throughout.
This method is widely used because it is simple to calculate and understand. Here is how it functions:

• The initial cost (or value) of the item is considered on the first day.
• A fixed amount is subtracted each year, representing the depreciation.

For this specific example:
- Initial Cost: \( \\(32,500 \)
- Depreciation Amount: \( \\)1950 \) annually
Understanding linear depreciation helps businesses budget and forecast effectively.

Graphing Linear Equations

Graphing linear equations involves plotting a straight line on a coordinate plane. For linear functions like the one in our exercise, the equation \( V(t) = 32500 - 1950t \) is represented.
In graphing, we use:

• "t" which is our x-variable (or time in years), and
• \( V(t) \) which is our y-variable (or value of the machinery).

To create a graph from this equation, follow these steps:

• Identify key points like the initial value (\( t = 0, V(t) = 32,500 \)), and after specific years, such as 5 years and 8 years, as calculated.
• Plot these points on the graph.
• Draw a line through these points; the line will slope downwards because the value diminishes over time.

This line provides a visual representation, making it easier to predict future values or assess how quickly the machinery's value approaches zero.

Solving Linear Equations

Solving linear equations is a crucial skill that helps us find unknown values. In our exercise, solving the linear equation \( V(t) = 32500 - 1950t \) allows us to determine when the machinery will have zero value.
Let's recall the equation for zero value:- \( 0 = 32500 - 1950t \) . By solving for \( t \), you get:

• Re-arrange to find \( t \): \( 1950t = 32500 \)
• Divide both sides by 1950: \( t = \frac{32500}{1950} \)
• Resulting in \( t \approx 16.67 \)

This tells us that it takes approximately 16.67 years for the machinery's value to decrease to zero.
Solving such equations provides concrete answers, offering insights into planning for asset replacement or other financial decisions.