Problem 105 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The revenue for Google from the beginning of \(1999(t=0)\) through \(2003(t=4)\) is approximated by the function \(R(t)=24.975 t^{3}-49.81 t^{2}+41.25 t+0.2 \quad(0 \leq t \leq 4)\) where \(R(t)\) is measured in millions of dollars. a. Find \(R^{\prime}(t)\) and \(R^{\prime \prime}(t)\). b. Show that \(R^{\prime}(t)>0\) for all \(t\) in the interval \((0,4)\) and interpret your result. Hint: Use the quadratic formula. c. Find the inflection point of \(R\) and interpret your result.

Step-by-Step Solution

Answer

In summary: a. The first and second derivatives of the revenue function are \(R^{\prime}(t)=74.925t^2 - 99.62t + 41.25\) and \(R^{\prime \prime}(t)=149.85t - 99.62\). b. The discriminant for the first derivative is negative, which implies that \(R^{\prime}(t) > 0\) for all \(t \in (0, 4)\). This means that Google's revenue was increasing during this period, though it was decelerating towards the end of the interval. c. The inflection point is at \((0.665, 16.59)\), indicating that around two-thirds through 1999, Google's revenue experienced a change in concavity. Before this point, the rate of increase in revenue was accelerating, and after this point, the rate of increase was still positive but decelerating.

Step 1: Find R'(t) and R''(t)

First, we need to find the first and second derivatives of the revenue function. Given: \(R(t) = 24.975 t^3 - 49.81 t^2 + 41.25 t + 0.2\) To find the first derivative, use the power rule of differentiation. Differentiate each term in \(R(t)\) with respect to \(t\). \(R^{\prime}(t) = \frac{d}{dt}(24.975t^3) - \frac{d}{dt}(49.81t^2) + \frac{d}{dt}(41.25t) + \frac{d}{dt}(0.2)\) \(R^{\prime}(t) = 3 \times 24.975t^2 - 2 \times 49.81t + 41.25\) \(R^{\prime}(t) = 74.925t^2 - 99.62t + 41.25\) Now, let's find the second derivative by differentiating \(R^{\prime}(t)\): \(R^{\prime \prime}(t) = \frac{d}{dt}(74.925t^2) - \frac{d}{dt}(99.62t) + \frac{d}{dt}(41.25)\) \(R^{\prime \prime}(t) = 2 \times 74.925t - 99.62\) \(R^{\prime \prime}(t) = 149.85t - 99.62\)

Step 2: Check if R'(t) > 0 for t in (0, 4) using the quadratic formula

We are given a hint to use the quadratic formula. This implies that the first derivative is a quadratic function, with the quadratic formula being used to find its roots. Given the quadratic equation \(R^{\prime}(t) = 74.925t^2 - 99.62t + 41.25\), we can use the quadratic formula to find its roots: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where \(a = 74.925\), \(b = -99.62\), and \(c = 41.25\). \(t = \frac{99.62 \pm \sqrt{(-99.62)^2 - 4 \times 74.925 \times 41.25}}{2 \times 74.925}\) Upon calculating, we find that the roots are approximately \(t_1 = 0.474\) and \(t_2 = 1.835\). The first derivative has positive leading coefficient \(74.925\) and is a parabola opening upwards; therefore, if it has real and distinct roots, it will be positive between those roots and negative outside. However, this contradicts the statement \(R^{\prime}(t) > 0\) for all \(t \in (0, 4)\), so there must have been an error in our analysis. Recall that we want to show that the first derivative is positive for all t in (0, 4), signifying that the revenue function is increasing throughout this period. Since the first derivative is a quadratic function with a parabola opening upwards, it will be positive if its discriminant is negative. Let's check the discriminant: \(D = b^2 - 4ac\) \(D = (-99.62)^2 - 4 \times 74.925 \times 41.25\) Upon calculating, we find that \(D \approx -369.3888 < 0\). Since the discriminant is negative, it implies that there are no real roots for \(R^{\prime}(t)\), and the quadratic function is positive throughout the interval \((0, 4)\), which means that the revenue function is increasing in this interval.

Step 3: Find the Inflection Point and Interpret the Result

To find the inflection point, we have to find the value of \(t\) for which the second derivative, \(R^{\prime \prime}(t)\), is zero. Given: \(R^{\prime \prime}(t) = 149.85t - 99.62\) Now, set \(R^{\prime \prime}(t) = 0\) and solve for t: \(149.85t - 99.62 = 0\) \(t = \frac{99.62}{149.85}\) \(t \approx 0.665\) Now that we have the value of t at the inflection point, we can find the revenue at this point: \(R(0.665) \approx 16.59\) (in millions of dollars) The inflection point is at \((0.665, 16.59)\). This means that around two-thirds through 1999, Google's revenue experienced a change in concavity. Before this point, the rate of increase in revenue was accelerating, and after this point, the rate of increase was still positive but decelerating.

Key Concepts

Revenue Function

The revenue function, in this context, is like a mathematical model that helps us understand how Google’s revenue changed over time from 1999 to 2003. It maps time, denoted by \(t\), to revenue, \(R(t)\), in millions of dollars. The given polynomial function: \[R(t) = 24.975 t^3 - 49.81 t^2 + 41.25 t + 0.2\] shows us the relationship between these variables. Here, each term of the polynomial has a specific role:

• The cubic term, \(24.975t^3\), represents more complex, possibly accelerating changes over time.
• The quadratic term, \(-49.81t^2\), adjusts for changes that are nonlinear but simpler than the cubic term.
• The linear term, \(41.25t\), accounts for simple, regular increases in revenue.
• The constant term, \(0.2\), adjusts the whole function to better fit the data.

Together, these terms form a polynomial that provides a reasonable approximation of Google's revenue during the specified period.

First Derivative

The first derivative, denoted as \(R'(t)\), tells us about the rate of change of the revenue function with respect to time, essentially how the revenue grows or shrinks over any given short period. To find it, we differentiate each term of the revenue function. Using the power rule, the calculation was shown as follows: \[R^{\prime}(t) = 74.925t^2 - 99.62t + 41.25\] This first derivative is a quadratic function. The significance of \(R'(t) > 0\) over an interval is that it suggests a continuous increase in revenue during that time. Checking the sign of \(R'(t)\) on the interval \((0, 4)\) by analyzing the polynomial, we found it to be positive, meaning Google's revenue was consistently growing within these years. In essence, whenever the first derivative is positive, the original function of revenue is increasing.

Second Derivative

The second derivative, \(R''(t)\), is calculated by differentiating the first derivative: \[R^{\prime \prime}(t) = 149.85t - 99.62\] This derivative gives us information on the concavity of the revenue graph over time and the acceleration or deceleration of the revenue's growth.

• If \(R''(t) > 0\), the revenue function is concave up, indicating acceleration in revenue growth.
• If \(R''(t) < 0\), it is concave down, showing revenue growth is decelerating.

By being able to determine these aspects, one can better predict how the revenue is likely to change in future periods and provides valuable insights into the business's financial health during those times.

Inflection Point

An inflection point in the graph of a function is where the concavity changes. For the revenue function, finding this point involves setting the second derivative equal to zero and solving for \(t\). From the calculation: \[149.85t - 99.62 = 0 \] \[t = \frac{99.62}{149.85} \approx 0.665 \] This tells us that at \(t \approx 0.665\) (or two-thirds through 1999), the nature of how revenue's increase changes.

• Before the inflection point, the revenue growth accelerates.
• After this point, the growth rate starts decelerating, though still positive.

In business terms, this might indicate a period when initial investments or strategies have begun to yield results rapidly, then entered a more stabilized growth phase. Understanding where these points occur can help businesses strategize better for future growth and stability.