Problem 109 - College Algebra with Modeling and Visualization

Question

The cost of tuition is directly proportional to the number of credits taken. If 11 credits cost \(\$ 720.50,\) find the cost of taking 16 credits. What is the constant of proportionality?

Step-by-Step Solution

Answer

16 credits cost \( \$1048.00 \), and the constant is \( 65.50 \).

Step 1: Understanding Direct Proportionality

When the cost is directly proportional to the number of credits, we have the relationship \( C = k \cdot n \), where \( C \) is the cost, \( k \) is the constant of proportionality, and \( n \) is the number of credits.

Step 2: Finding the Constant of Proportionality

From the given information, 11 credits cost \( \$720.50 \). Use the formula \( C = k \cdot n \) and substitute \( C = 720.50 \) and \( n = 11 \). So, \( 720.50 = k \cdot 11 \). Solve for \( k \) by dividing both sides by 11: \( k = \frac{720.50}{11} \).

Step 3: Calculate the Constant 'k'

Perform the division to find \( k \). \( k = \frac{720.50}{11} = 65.50 \). The constant of proportionality is \( 65.50 \).

Step 4: Calculating Cost for 16 Credits

Use the formula \( C = k \cdot n \) with \( k = 65.50 \) and \( n = 16 \) to find the cost for 16 credits. Substitute the values: \( C = 65.50 \cdot 16 \).

Step 5: Final Calculation

Multiply to find the total cost: \( C = 65.50 \times 16 = 1048.00 \). Thus, the cost for 16 credits is \( \$1048.00 \).

Key Concepts

Constant of Proportionality

In mathematics and real-world applications, the constant of proportionality is a crucial concept when dealing with direct proportionality. It represents a fixed multiplier that links two variables where a change in one leads to a proportional change in the other. This is expressed in the formula \( C = k \cdot n \), where \( C \) is the total cost, \( k \) is the constant of proportionality, and \( n \) is the number of credits.
To find the constant of proportionality, you need data where one of the variables (like cost) and its corresponding value (like number of credits) are known. For example, the data in our exercise gives us that 11 credits cost \( \\(720.50 \). Plugging these values into our formula, \( 720.50 = k \cdot 11 \), lets us solve for \( k \): \( k = \frac{720.50}{11} \). Performing this calculation gives \( k = 65.50 \).
This constant value, \( 65.50 \), essentially tells us that each credit costs \( \\)65.50 \). No matter how many credits you take, multiplying by \( 65.50 \) gives you the total cost.

Algebraic Equations

Algebraic equations are a way of representing mathematical relationships using variables and constants. In our exercise, the algebraic equation \( C = k \cdot n \) becomes a tool for modeling the relationship between the cost of tuition and the number of credits taken. This equation is a simple linear model showing direct proportionality.
To manipulate algebraic equations effectively, understanding the role of each component is crucial:

• \( C \): The total cost, which changes depending on how many credits are taken.
• \( k \): The constant of proportionality, representing cost per credit.
• \( n \): The number of credits, which determines the overall cost.

Solving such equations often involves isolating one variable to find its value. As shown in our steps, after substituting the known values, we divided to find \( k \). Likewise, to find the total cost for a different number of credits, the same process is applied using the calculated \( k \). Understanding this structure empowers students to solve similar problems in different contexts.

Cost Calculation

Cost calculation with direct proportion involves using a constant rate to multiply by a quantity, resulting in a total cost. In our example, once the constant of proportionality \( k \) is known to be \( 65.50 \), calculating the cost for a new number of credits is straightforward.
To calculate the cost for 16 credits, you substitute \( k = 65.50 \) and \( n = 16 \) into the formula \( C = k \cdot n \). Doing this multiplication, \( C = 65.50 \times 16 \), results in \( \$1048.00 \).
This approach simplifies the process of calculating costs because it relies on a standard rate per unit, making it easily adaptable to different scenarios. For any number of credits, you multiply by the constant to find the total cost quickly. This clear and structured method helps not only in academic exercises but also in practical applications like budgeting or financial planning.