Problem 63 - Brief Applied Calculus

Question

63-64. GENERAL: College Tuition The following graph shows the average annual college tuition costs (tuition and fees) for a year at a private nonprofit or public fouryear college. The data are given for five-year intervals. The tuition for a private college is approximated by the function \(f(x)=650 x^{2}+3000 x+12,000\), where \(x\) is the number of five-year intervals since the academic year \(1995-96\) (so the years in the graph are numbered \(x=0\) through \(x=3\) ). a. Use this function to predict tuition in the academic year 2020-21. [Hint: What \(x\) -value corresponds to that year?] b. Find the derivative of this function for the \(x\) -value that you used in part (a) and interpret it as a rate of change in the proper units. c. From your answer to part (b), estimate how rapidly tuition will be increasing per year in \(2020-21\).

Step-by-Step Solution

Answer

The predicted tuition for 2020-21 is \(\$43,250\). Tuition increases annually by \(\$1,900\) in 2020-21.

Step 1: Understand the Function and Year Conversion

The given function is \( f(x) = 650x^2 + 3000x + 12,000 \), where \( x \) represents the number of five-year intervals since the academic year 1995-96. We need to express the year 2020-21 in terms of \( x \). Since 1995-96 corresponds to \( x = 0 \), each five-year interval increments \( x \) by 1. Therefore, to find \( x \) for 2020-21: \[ x = \frac{2021 - 1995}{5} = 5.2. \] However, since \( x \) must be a whole number, we approximate to the nearest valid interval, giving us \( x = 5 \).

Step 2: Predict Tuition for 2020-21

Substitute \( x = 5 \) into the function \( f(x) = 650x^2 + 3000x + 12,000 \) to find the predicted tuition for 2020-21: \[ f(5) = 650(5)^2 + 3000(5) + 12,000 = 650(25) + 15000 + 12000. \] Calculate: \( 650 \times 25 = 16,250 \). Sum the terms to get \( f(5) = 16,250 + 15,000 + 12,000 = 43,250 \). Therefore, the predicted tuition is \($43,250\).

Step 3: Find the Derivative of the Function

To find the rate of change, calculate the derivative of the function \( f(x) = 650x^2 + 3000x + 12,000 \). Use the power rule: \( f'(x) = 2 \times 650x + 3000 = 1300x + 3000 \).

Step 4: Evaluate the Derivative at x = 5

Substitute \( x = 5 \) into the derivative to find the rate of change: \( f'(5) = 1300(5) + 3000 = 6500 + 3000 = 9500 \). This indicates the rate of change in tuition is \( \$9,500 \) per five-year interval in 2020-21.

Step 5: Estimate Annual Rate of Increase in Tuition

Since the rate of change is \\(9,500 per five-year interval, divide by 5 to find the annual rate: \[ \text{Annual Rate} = \frac{9500}{5} = 1900. \] Therefore, tuition is increasing at approximately \( \\)1,900 \) per year in 2020-21.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving variables raised to whole number powers, combined using addition, subtraction, and multiplication. In our tuition prediction problem, the polynomial function is given by \( f(x) = 650x^2 + 3000x + 12,000 \). This function represents the average college tuition based on five-year intervals since 1995-96, where \( x \) is the interval count.
Here, each term of the function plays a specific role:

• \( 650x^2 \): The quadratic term, which suggests that the tuition cost increases quadratically over time.
• \( 3000x \): The linear term signifies a consistent rate of tuition increase.
• \( 12,000 \): The constant term indicates the base cost of tuition in the starting year (1995-96).

These components together provide a comprehensive mathematical model for predicting tuition costs. Understanding polynomial functions is crucial for analyzing and predicting real-world economic factors such as these.

Derivative Calculation

A derivative provides a mathematical way to determine how a function is changing at any given point. In the context of our tuition cost function, the derivative tells us how fast tuition is expected to rise as the years go by. The derivative of a polynomial function like \( f(x) = 650x^2 + 3000x + 12,000 \) can be calculated using the power rule, which involves differentiating each term separately.
For our function:

• The derivative of \( 650x^2 \) using the power rule (multiply the exponent by the coefficient and reduce the exponent by one) is \( 1300x \).
• The derivative of \( 3000x \) is simply the coefficient \( 3000 \) because the derivative of \( x \) is 1.
• The derivative of a constant \( 12,000 \) is zero since constants do not change.

Adding these, the overall derivative of the function is \( f'(x) = 1300x + 3000 \). Evaluating this derivative at \( x = 5 \), we find \( f'(5) = 9500 \), suggesting tuition rises at \( \$9500 \) per five-year interval around the 2020-21 academic year.

Rate of Change in Economics

Understanding the rate of change is vital for economic forecasting and decision-making. In economics, it tells us about the speed at which a given quantity like tuition costs is changing over a certain period. Predicting tuition increases allows institutions and students to plan their finances more effectively.

In our context, calculating the derivative gives the rate of change in tuition per five-year interval, which was found to be \( \\(9500 \). To understand its effect annually, divide by the number of years in an interval. Thus, we conclude the tuition cost increases by about \( \\)1900 \) per year around 2020-21.
This simple derivative calculation offers a snapshot of how tuition fees might trend based on historical patterns, highlighting the importance of calculus in economic decision-making and planning.