Problem 66 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

EfFECT OF ADVERTISING ON SALES The relationship between the amount of money \(x\) that Cannon Precision Instruments spends on advertising and the company's total sales \(S(x)\) is given by the function \(S(x)=-0.002 x^{3}+0.6 x^{2}+x+500 \quad(0 \leq x \leq 200)\) where \(x\) is measured in thousands of dollars. Find the rate of change of the sales with respect to the amount of money spent on advertising. Are Cannon's total sales increasing at a faster rate when the amount of money spent on advertising is (a) \(\$ 100,000\) or (b) \(\$ 150,000\) ?

Step-by-Step Solution

Answer

The rate of change of sales with respect to the advertising spending is given by the derivative \(S'(x) = -0.006x^2 + 1.2x + 1\). When \(x = 100\), the rate of change is \(61\), and when \(x = 150\), the rate of change is \(46\). Therefore, Cannon's total sales are increasing at a faster rate when the amount of money spent on advertising is \(\$100,000\).

Step 1: Find the derivative of the function S(x)

To find the rate of change of sales with respect to the money spent on advertising, we find the derivative of S(x) with respect to x. Given, \(S(x) = -0.002x^3 + 0.6x^2 + x + 500\) Differentiate S(x) with respect to x: \(S'(x) = \frac{dS}{dx}( -0.002x^3 + 0.6x^2 + x + 500)\) Using the power rule, we get: \(S'(x) = -0.006x^2 + 1.2x + 1\)

Step 2: Evaluate the derivative at x = 100 and x = 150

Now, we'll plug in the values of x (100 and 150) into the derivative to find the rate of change at these points. For x = 100: \(S'(100) = -0.006(100)^2 + 1.2(100) + 1\) \(S'(100) = -0.006(10000) + 120 + 1\) \(S'(100) = -60 + 120 + 1\) \(S'(100) = 61\) For x = 150: \(S'(150) = -0.006(150)^2 + 1.2(150) + 1\) \(S'(150) = -0.006(22500) + 180 + 1\) \(S'(150) = -135 + 180 + 1\) \(S'(150) = 46\)

Step 3: Compare the rate of change at x = 100 and x = 150

We found that the rate of change at x = 100 is 61 and at x = 150 is 46. Since 61 > 46, we can conclude that Cannon's total sales are increasing at a faster rate when the amount of money spent on advertising is \(\$100,000\).

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It is a core concept that aids in understanding how functions behave. To compute the derivative of a function, we typically use rules like the power rule, product rule, or quotient rule.

• For example, the power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
• In this exercise, applying the power rule on each term of the polynomial \( S(x) \) gives the derivative \( S'(x) = -0.006x^2 + 1.2x + 1 \).

The derivative \( S'(x) \) provides us insight into how the company's sales might change as advertising spending changes. This is invaluable for businesses analyzing the effectiveness of their advertising budgets.

Rate of Change

The rate of change in calculus is fundamentally about how one quantity changes in relation to another. In this context, it refers to how the company's total sales \( S(x) \) change with respect to advertising spend \( x \). This is commonly assessed using derivatives.

• If \( S'(x) > 0 \), it indicates sales are increasing as advertising increases.
• If \( S'(x) < 0 \), sales are decreasing as the advertising expenditure increases.
• Here, the focus is on two specific points: \( x = 100 \) and \( x = 150 \).

Calculating \( S'(100) \) and \( S'(150) \) helps determine if the sales grow faster at one level of advertising spend compared to another. Understanding these changes enables businesses to optimize spending for maximum growth impact.

Advertising Impact on Sales

Advertising can significantly impact sales, and understanding this relationship is crucial for businesses. Through calculus, specifically by using derivatives, we can quantify this impact.
In the exercise, the company knows its sales model: \( S(x) = -0.002x^3 + 0.6x^2 + x + 500 \). This tells us that there's a complex relationship between advertising spend \( x \) and sales.

• A positive derivative value like \( S'(100) = 61 \) shows that with every additional thousand dollars spent at this point, sales increase by roughly 61 units (in thousands).
• When \( S'(150) = 46 \), it shows a slower growth rate compared to when spending was at \( 100 \), implying diminishing returns as more is spent.

Businesses can use such insights to decide their advertising budget allocations effectively and strategically.

Polynomial Functions

Polynomial functions are mathematical expressions involving variables raised to whole number powers. They are among the simplest forms of functions and commonly appear in real-world applications like finance and economics.

• The given function \( S(x) = -0.002x^3 + 0.6x^2 + x + 500 \) is a polynomial of degree 3, also referred to as a cubic polynomial.
• Such functions are versatile for modeling complex relationships, as seen with Cannon's sales prediction.

Calculating derivatives of polynomial functions is straightforward due to their specific structure, enabling quick assessments of changes in predicted outcomes, which is essential for business planning and analysis. Understanding polynomial functions and their derivatives helps in effectively forecasting sales, expenses, and profits.