Problem 68 - Brief Applied Calculus

Question

BUSINESS: CD Sales Suppose that after \(x\) months, monthly sales of a compact disc are predicted to be \(S(x)=x^{2}\left(8-x^{3}\right)\) thousand (for \(0 \leq x \leq 2\) ). Find the rate of change of the sales after 1 month.

Step-by-Step Solution

Answer

The rate of change of sales after 1 month is 11 thousand CDs/month.

Step 1: Understand the Given Function

The sales function given is \( S(x) = x^2 (8 - x^3) \). It models the sales of CDs in thousands after \( x \) months. We want to find the rate of change of sales when \( x = 1 \).

Step 2: Differentiate the Sales Function

To find the rate of change of sales, we need the derivative of the function \( S(x) \). Using the product rule, \((f g)' = f' g + f g'\), we differentiate: Let \( f(x) = x^2 \) and \( g(x) = 8 - x^3 \).Then, \( f'(x) = 2x \) and \( g'(x) = -3x^2 \). The derivative of \( S(x) \) is:\[S'(x) = f'(x) g(x) + f(x) g'(x) = 2x(8 - x^3) + x^2(-3x^2).\]

Step 3: Simplify the Derivative Expression

Substitute the expressions into the derivative:\[S'(x) = 2x(8 - x^3) + x^2(-3x^2) = 16x - 2x^4 - 3x^4 = 16x - 5x^4.\]

Step 4: Evaluate the Derivative at x = 1

To find the rate of change of sales after 1 month, substitute \( x = 1 \) into the derivative:\[S'(1) = 16(1) - 5(1)^4 = 16 - 5 = 11.\]

Step 5: Interpret the Result

The result \( S'(1) = 11 \) indicates that the rate of change of sales after 1 month is 11 thousand CDs per month. This means that, at the start of the second month, sales are increasing at a rate of 11 thousand CDs per month.

Key Concepts

product rule in calculus

When dealing with functions that are products of two simpler functions, like the sales function \( S(x) = x^2 (8 - x^3) \), the product rule in calculus helps find the derivative. This rule is essential for differentiation when one function is multiplied by another. The product rule formula is:

• \((f \cdot g)' = f' \cdot g + f \cdot g'\)

Here, \( f(x) = x^2 \) and \( g(x) = 8 - x^3 \). We derive each function separately:

• \( f'(x) = 2x \)
• \( g'(x) = -3x^2 \)

Using the product rule, the derivative of \( S(x) \) becomes:

• \( S'(x) = f'(x)g(x) + f(x)g'(x) \)
• \( S'(x) = 2x(8 - x^3) + x^2(-3x^2) \)

This process clearly shows how the product rule works and its importance for understanding complex functions in calculus.

rate of change

The rate of change in calculus illustrates how a quantity is changing over time or in response to something else. It gives us insights into the behavior of the function at a particular point.In this exercise, we are interested in finding the rate at which monthly CD sales are changing after one month. Calculus helps us find this by calculating the derivative of the sales function, \( S(x) \). The derivative at a particular point, \( S'(1) = 11 \), tells us the sales are increasing by 11 thousand CDs per month after one month.This provides a detailed understanding of whether the trend in CD sales is upwards or downwards. If the rate is a positive number, it indicates an increase, as seen here. This concept is not only fundamental in calculus but is also crucial in fields like physics and engineering where understanding change is imperative.

differentiation techniques

Differentiation techniques in calculus are methods used to compute the derivative of functions. These techniques allow us to find rates of change, which can be immensely useful in various applications.In the provided exercise, a specific technique known as the product rule was used. However, there are also other techniques such as:

• Power Rule: Used when a function is a power of \( x \). If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Chain Rule: Useful for composite functions, allowing us to differentiate a function composed inside another function.
• Quotient Rule: Used for functions that are divided by each other, offering a systematic way to find the derivative of a quotient.

These techniques provide a powerful set of tools for tackling a wide range of problems in calculus. Knowing when and how to apply each rule helps simplify complex differentiation tasks, leading to more efficient problem-solving in mathematics and its applications.