Problem 24 - College Algebra and Calculus: An Applied Approach

Question

You are given the 2005 value of a product and the rate at which the value is expected to change during the next 5 years. Use this information to write a linear equation that gives the dollar value of the product in terms of the year. (Let \(t=5\) represent 2005.) 2005 Value \(\quad\) Rate $$\$ 45,000 \quad \$ 2800$$ decrease per year

Step-by-Step Solution

Answer

The linear equation that gives the dollar value of the product in terms of the year since 2005 is \( y = - \$2800(t-5) + \$45000 \)

Step 1: Identify Key variables

The value of the product in 2005 is \( \$45,000. t=5 represents year 2005. The rate of decrease per year is \( \$2800.

Step 2: Set up the Linear Equation

The general structure of a linear equation is \( y = mx + c \). Here, \( y \) will be the value of the product, \( t \) will represent the year, \( m \) is the rate of decrease and \( c \) is the initial value of the product in 2005.

Step 3: Substitute the values into the linear equation

Substitute \( c = \$45,000 \), and \( m = - \$2800 \) into the linear equation to get \( y = - \$2800t + \$45000 \).

Step 4: Adjust the equation for 2005 as t=5

Since we are given that \( t=5 \) represents 2005, we can change our equation to reflect the number of years since 2005. The new equation will thus be \( y = - \$2800(t-5) + \$45000 \)

Key Concepts

Rate of Change

In the context of linear equations, the rate of change is an essential concept that refers to how a quantity changes in relation to another. Here, it's the decline in the product's value over time. It's represented by the coefficient in a linear equation. In our exercise, the rate of change is \\(2800\ each year.
Rate of change can be positive or negative. It tells us whether the quantity increases or decreases as one variable increases.
- **Positive rate of change:** The quantity increases as the independent variable (i.e., time) increases.
- **Negative rate of change:** Indicated by the exercise’s decrement of \\)2800\ per year, meaning the product loses value annually.
Understanding the rate of change helps you describe the behavior of a linear function and make predictions about future values.

Linear Function

A linear function is a mathematical equation that forms a straight line when graphed. It's expressed in the form \ y = mx + c\, where:
- **\m\** is the slope or rate of change. It tells us how steeply the line increases or decreases.
- **\c\** is the y-intercept, which shows the starting point of the line on the y-axis.
In this exercise, the linear function is \(y = -2800t + 45000\). It indicates that the value of a product decreases by \\(2800\ every year starting from \\)45000\ in 2005.
Linear functions are straightforward and predict future values easily. By understanding them, you can plot future points on the graph, showing how the product value will trend over the years.

Algebraic Modeling

Algebraic modeling involves creating algebraic expressions or equations that represent real-life scenarios. Here, it allows us to create a model predicting the decrease in product value over several years.
In this task, algebraic modeling results in the equation \( y = -2800(t-5) + 45000\).
- **Visualize real scenarios:** Producing a model helps to visualize changes over time or other conditions.
- **Make informed decisions:** You have the power to predict outcomes and make decisions based on the algebraic model, such as understanding how many years it will take for the product value to reach a certain amount.
Through algebraic modeling, we can reliably predict that the product's value decreases consistently each year. It serves as a practical tool for planning and forecasting in various contexts.