Problem 68 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The average annual price of single-family homes in Massachusetts between 1990 and 2002 is approximated by the function \(P(t)=-0.183 t^{3}+4.65 t^{2}-17.3 t+200 \quad(0 \leq t \leq 12)\) where \(P(t)\) is measured in thousands of dollars and \(t\) is measured in years, with \(t=0\) corresponding to 1990 . In what year was the average annual price of single-family homes in Massachusetts lowest? What was the approximate lowest average annual price? Hint: Use the quadratic formula.

Step-by-Step Solution

Answer

The lowest average annual price of single-family homes in Massachusetts was in the year 1991, with the approximate lowest average annual price being $186,860.

Step 1: Determine the first derivative of the polynomial function

Find the first derivative of the given function \(P(t) = -0.183t^3 + 4.65t^2 -17.3t + 200\) by applying the power rule of differentiation: \[ \frac{dP}{dt} = -3(0.183)t^2 + 2(4.65)t - 17.3 \] Simplify the derivative: \(\frac{dP}{dt} = -0.549t^2 + 9.3t - 17.3\)

Step 2: Find the critical points of the function

Set the derivative equal to zero to find the critical points: \[ -0.549t^2 + 9.3t - 17.3 = 0 \]

Step 3: Apply quadratic formula to solve for t

Use the quadratic formula to solve the equation: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -0.549\), \(b = 9.3\), and \(c = -17.3\). \[ t = \frac{-9.3 \pm \sqrt{9.3^2 - 4(-0.549)(-17.3)}}{2(-0.549)} \] After solving, we get the two possible values of \(t\): \(t_1\approx 1.22\) and \(t_2\approx 25.55\). However, since the given timeframe is from \(0\) to \(12\) years, only \(t_1\) falls into this range. Therefore, the minimum value occurs at \(t = 1.22\).

Step 4: Find the corresponding year

To find the year in which the lowest average annual price occurred, we add the value of \(t\) to the year 1990: Year \(= 1990 + 1.22\) Year \(\approx \) 1991

Step 5: Calculate the lowest average annual price

Now that we have the value of \(t\) that corresponds to the lowest point, we can plug that value back into the original function to find the lowest average annual price: \[ P(1.22) = -0.183(1.22)^3 + 4.65(1.22)^2 - 17.3(1.22) + 200 \] After calculating, we find that \(P(1.22) \approx 186.86\). Since the price is measured in thousands of dollars, the lowest average annual price is approximately $186,860. In conclusion, the lowest average annual price of single-family homes in Massachusetts occurred in the year 1991, with an approximate value of $186,860.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving variables raised to different powers, combined with coefficients. These functions can take on different shapes based on the degree and coefficients. In the exercise, the function given is a cubic polynomial: \[ P(t) = -0.183t^3 + 4.65t^2 - 17.3t + 200 \]

• The degree of this polynomial is 3, given by the highest power of the variable \( t \).
• Polynomial functions can model complex real-world phenomena, like the fluctuations in housing prices over time.

Polynomial functions are particularly useful in calculus and algebra to predict trends and find key features such as peaks or troughs in the data.​ Break down each part of the polynomial to understand its impact: the coefficient of \(t^3\) affects the curvature, \(t^2\) influences the rise and fall, and the linear and constant terms adjust the starting point and slope. Each component provides insight into the behavior of the function.

Derivatives

In calculus, the derivative of a function measures the rate at which the function's value changes with respect to change in its variable. This concept is crucial for understanding polynomial functions. For the given function, the derivative is calculated by applying the power rule: \[ \frac{dP}{dt} = -3(0.183)t^2 + 2(4.65)t - 17.3 \] Which simplifies to: \[ \frac{dP}{dt} = -0.549t^2 + 9.3t - 17.3 \]

• The power rule helps to find the derivative of polynomials by multiplying the exponent by the coefficient and reducing the exponent by one.
• This specific derivative tells us how the price is changing over time.

Finding the derivative filters out static components of the polynomial, leaving a formula that describes how the function's rate of change is affected by the variable \(t\). This is key to identifying points where the function's increase or decrease momentarily stops.

Quadratic Formula

The quadratic formula is a fundamental tool in algebra, used to solve equations where the highest term is squared. In this exercise, we solve the equation resulting from the derivative set to zero using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, for the expression \[ -0.549t^2 + 9.3t - 17.3 = 0 \]

• \(a = -0.549\), \(b = 9.3\), and \(c = -17.3\).
• Applying the formula, one can find the values of \(t\) where the derivative is zero.

These solutions for \(t\) represent the points in time where the rate of change (derivative) is zero, known as critical points. The quadratic formula simplifies the process of finding these critical points, allowing us to evaluate when a polynomial's rate of change is at a minimum or maximum.

Critical Points

Critical points occur where the derivative of a function equals zero or is undefined. For polynomials, this typically indicates a local minimum, maximum, or a point of inflection.

• These points are important for understanding where key changes in direction occur in the function's graph.
• For the housing price problem, the critical point \(t = 1.22\) represents a local minimum.

By evaluating the polynomial at these critical points, we can determine where the lowest or highest values occur, and gain insights into characteristics such as stability or volatility over time. Calculating the original function at these points confirms the particular value reflects a lowest price.