Problem 65 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The total worldwide box-office receipts for a long-running movie are approximated by the function $$ T(x)=\frac{120 x^{2}}{x^{2}+4} $$ where \(T(x)\) is measured in millions of dollars and \(x\) is the number of years since the movie's release. How fast are the total receipts changing \(1 \mathrm{yr}, 3 \mathrm{yr}\), and \(5 \mathrm{yr}\) after its release?

Step-by-Step Solution

Answer

The total receipts are changing at a rate of \(38.4\) million dollars per year after \(1\) year, \(17.045\) million dollars per year after \(3\) years, and \(5.711\) million dollars per year after \(5\) years.

Step 1: Calculate the derivative of the function T(x) with respect to x

To find the rate at which the total receipts are changing, we need to find the derivative of the given function, T(x), with respect to x. The function is given by: \[ T(x)=\frac{120 x^{2}}{x^{2}+4} \] To find the derivative, we can use the quotient rule, which states that if we have a function \[ h(x) = \frac{f(x)}{g(x)} \] then \[ h'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2} \] Applying this rule to our function T(x), we have: \[ \frac{d}{dx}T(x) = \frac{d}{dx}\bigg(\frac{120x^2}{x^2 + 4}\bigg) \] Let \(f(x) = 120x^2\) and \(g(x) = x^2 + 4\), then \(f'(x) = 240x\) and \(g'(x) = 2x\). Using the quotient rule, the derivative is: \[ T'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2} = \frac{(240x)(x^2 + 4) - (2x)(120x^2)}{(x^2 + 4)^2} \]

Step 2: Simplify the derivative

Now let's simplify T'(x): \[ T'(x) = \frac{240x^3 + 960x - 240x^3}{(x^2 + 4)^2} = \frac{960x}{(x^2 + 4)^2} \] So, the rate at which the total receipts are changing is: \[ T'(x) = \frac{960x}{(x^2 + 4)^2} \]

Step 3: Evaluate the derivative at x = 1, x = 3, and x = 5

Now we need to find the rate at which the total receipts are changing 1 year, 3 years, and 5 years after its release. We can do this by evaluating T'(x) at x = 1, x = 3, and x = 5. 1 year after release: \[ T'(1) = \frac{960(1)}{(1^2 + 4)^2} = \frac{960}{25} = 38.4 \] 3 years after release: \[ T'(3) = \frac{960(3)}{(3^2 + 4)^2} = \frac{2880}{169} \approx 17.045 \] 5 years after release: \[ T'(5) = \frac{960(5)}{(5^2 + 4)^2} = \frac{4800}{841} \approx 5.711 \] So, the total receipts are changing at a rate of 38.4 million dollars per year after 1 year, 17.045 million dollars per year after 3 years, and 5.711 million dollars per year after 5 years.

Key Concepts

Derivative

When we talk about derivatives in calculus, we are essentially discussing how a function changes as its input changes. More specifically, the derivative of a function at a certain point gives us the instantaneous rate of change at that point. Think of it as measuring how fast something is happening at a precise moment. In this exercise, we are looking at a function that models box-office receipts over time, and the derivative tells us how fast these receipts are increasing or decreasing. To find the derivative of a function like the one given, sometimes we need special rules. The function here is a quotient, meaning one function divided by another, which we calculate using a specific method known as the quotient rule.

Quotient Rule

The quotient rule is a technique used to find the derivative of a division (or quotient) of two functions. If you have two functions, say \( f(x) \) and \( g(x) \), and you're interested in the derivative of their quotient \( h(x) = \frac{f(x)}{g(x)} \), the quotient rule is your tool. The formula is:\[ h'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2}\]Here's the key:

• \( f'(x) \) is the derivative of the numerator.
• \( g(x) \) and \( g'(x) \) are the function and derivative of the denominator.
• The bottom of the fraction in the quotient rule is simply \( g(x)^2 \).

To visualize, imagine first multiplying the derivative of the top function by the bottom function, and then subtracting the result of multiplying the top function by the derivative of the bottom function. All divided by the square of the bottom function. Applying this rule to our exercise allows us to see the changing dynamics of the movie receipts as time progresses.

Rate of Change

The concept of rate of change is central to understanding derivatives. It represents how a quantity changes over a particular period of time. In practical terms, it's how fast something is increasing or decreasing. In our exercise, this is translated into how quickly the movie's total receipts change year by year.For a real-world application, consider how you observe changes in your bank account. If you deposit money, the rate of change is positive. If you withdraw money, it's negative. Similarly, in the exercise, the derivative \( T'(x) \) gives the rate of change of the total receipts. Evaluating \( T'(x) \) at specific years (1, 3, and 5) provides specific insights into how fast the movie is earning money during those times:

• At 1 year, the rate is 38.4 million dollars per year.
• At 3 years, it slows down to approximately 17.045 million.
• At 5 years, it further decreases to about 5.711 million.

This deceleration suggests that the movie is earning less as years go by, a common trend as initial excitement wanes.