Problem 69 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

After the economy softened, the sky-high office space rents of the late 1990 s started to come down to earth. The function \(R\) gives the approximate price per square foot in dollars, \(R(t)\), of prime space in Boston's Back Bay and Financial District from \(1997(t=0)\) through 2002, where \(R(t)=-0.711 t^{3}+3.76 t^{2}+0.2 t+36.5 \quad(0 \leq t \leq 5)\) Show that the office space rents peaked at about the middle of 2000 . What was the highest office space rent during the period in question? Hint: Use the quadratic formula.

Step-by-Step Solution

Answer

The office space rents peaked at about the middle of 2000 (t=3.12). The highest office space rent during the period in question was approximately $39.88 per square foot.

Step 1: Find the derivative of R(t)

We'll find the derivative of \(R(t) = -0.711t^3 + 3.76t^2 + 0.2t + 36.5\). Using power rule for derivatives, the derivative will be: \(R'(t) = \dfrac{dR}{dt} = -3(0.711) t^2 + 2(3.76) t + 0.2\) Now we should simplify the expression: \(R'(t) = -2.133t^2 + 7.52t + 0.2\)

Step 2: Find the critical points

A critical point is where the derivative equals 0 or does not exist. In our case, the derivative exists for all values of t, so we only need to find the values of t when \(R'(t) = 0\). We have a quadratic equation, so we can use the quadratic formula to solve it: \(t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where \(a = -2.133, b = 7.52, c = 0.2\).

Step 3: Calculate the critical points

Now, we can use the quadratic formula with the given values: \(t = \dfrac{-7.52 \pm \sqrt{(7.52)^2 - 4(-2.133)(0.2)}}{2(-2.133)}\) Calculate the value within the square root, \(\Delta = (7.52)^2 - 4(-2.133)(0.2) = 56.5504 + 1.7064 = 58.2568\) Now we use the values in the formula: \(t = \dfrac{-7.52 \pm \sqrt{58.2568}}{-4.266}\) Now we have two possible values for t: \(t_1 = \dfrac{-7.52 + \sqrt{58.2568}}{-4.266} \approx 3.12\) \(t_2 = \dfrac{-7.52 - \sqrt{58.2568}}{-4.266}\) (This is not in the interval \([0, 5]\), so we ignore this value.) Since \(t_1 = 3.12\) is within our interval of \([0, 5]\), we should check if this is the maximum by comparing the rent values at the endpoints and the critical point.

Step 4: Evaluate the rent values at critical points and endpoints

Now, let's evaluate \(R(t)\) at the three points we need to check - \(t=0,3.12,5\). \(R(0) = -0.711(0)^3 + 3.76(0)^2 + 0.2(0) + 36.5 = 36.5\) \(R(3.12) = -0.711(3.12)^3 + 3.76(3.12)^2 + 0.2(3.12) + 36.5 \approx 39.88\) \(R(5) = -0.711(5)^3 + 3.76(5)^2 + 0.2(5) + 36.5 = 31\)

Step 5: Compare the rent values and find the maximum

Comparing these values: \(R(0) = 36.5\) \(R(3.12) \approx 39.88\) \(R(5) = 31\) We can see that the highest rent value is at \(t = 3.12\), which corresponds to the middle of 2000. The maximum rent during this period was approximately $39.88 per square foot.

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It's essentially the function's "slope" or "instantaneous rate of change". For a function expressed as

• \( y = f(x) \)

the derivative is expressed as

• \( \frac{dy}{dx} \)

This notation highlights how the derivative shows how small changes in \(x\) affect \(y\). We can think of the derivative as the tool that helps us understand the function's behavior in a neighborhood around a given point.
In our original problem, we found the derivative of the function \( R(t) = -0.711t^3 + 3.76t^2 + 0.2t + 36.5 \). Using the power rule, which states that \( \frac{d}{dx}[x^n] = nx^{n-1} \), we compute the derivative:

• \( R'(t) = -2.133t^2 + 7.52t + 0.2 \)

This derivative gives us the rate of change of rents over time, pinpointing places where rents may have peaked or bottomed out.

Quadratic Formula

The quadratic formula is a crucial mathematical tool used to find the roots of a quadratic equation of the form

• \( ax^2 + bx + c = 0 \)

By applying the formula

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

we can determine where the equation's graph crosses the x-axis, or in practical terms, solve for when a given parabolic event (like rent increases) reaches a specific value. In many problems, these solutions guide us to critical points that may represent maximum or minimum values.
In our exercise, the derivative function \( R'(t) \) is set to zero to locate critical points. To solve

• \( -2.133t^2 + 7.52t + 0.2 = 0 \)

we used the quadratic formula to find values of \(t\), which eventually led to \( t_1 = 3.12 \) as the viable critical point within the considered timeframe. This value helped us to investigate whether it mirrors a peak rent value.

Critical Points

Critical points are important when analyzing the behavior of a function, especially to determine points of interest such as peaks, valleys, or plateaus. A critical point occurs when the derivative equals zero or is undefined. It is at these points where we can find potential local maxima or minima in a function.
In the context of our exercise, after finding the derivative,

• \( R'(t) = -2.133t^2 + 7.52t + 0.2 \)

we set it equal to zero to locate the critical points. Solving this using the quadratic formula revealed a significant point \( t = 3.12 \). Such critical points are then examined to determine whether they correspond to local maxima, minima, or inflection points, which ultimately help us understand the changes in rent dynamics.

Maxima and Minima

Maxima and minima are used to describe the highest and lowest points of a function on a specific interval, respectively. These points tell us where a function's output reaches its peak or trough. To find these, we evaluate the function at the critical points and boundaries of the interval.
For the rent equation over the interval

• \( 0 \leq t \leq 5 \)

we calculated \( R(t) \) at critical point \( t = 3.12 \), alongside the interval boundaries, \( t = 0 \) and \( t = 5 \). This step involved assessing:

• \( R(0) = 36.5 \)
• \( R(3.12) \approx 39.88 \)
• \( R(5) = 31 \)

By comparing these values, we identified \( t = 3.12 \) as the point of maximum rent, indicating the rent peaked around the middle of 2000 at approximately \$39.88. Understanding these maxima and minima is pivotal for analyzing financial trends, like the fluctuating office rents in this exercise.