Problem 67 - Precalculus Mathematics for Calculus

Question

The effectiveness of a television commercial depends on how many times a viewer watches it. After some experiments an advertising agency found that if the effectiveness \(E\) is measured on a scale of 0 to \(10,\) then $$E(n)=\frac{2}{3} n-\frac{1}{90} n^{2}$$ where \(n\) is the number of times a viewer watches a given commercial. For a commercial to have maximum effectiveness, how many times should a viewer watch it?

Step-by-Step Solution

Answer

A viewer should watch the commercial 30 times for maximum effectiveness.

Step 1: Understand the Problem

The effectiveness of a commercial is given by the function \( E(n) = \frac{2}{3}n - \frac{1}{90}n^2 \). We need to find the value of \( n \) that maximizes this function.

Step 2: Differentiate to Find Critical Points

To find the value of \( n \) that maximizes \( E(n) \), we take the first derivative of \( E(n) \). We have:\[ E'(n) = \frac{d}{dn} \left( \frac{2}{3}n - \frac{1}{90}n^2 \right) = \frac{2}{3} - \frac{2}{90}n. \]

Step 3: Set Derivative to Zero

Set the first derivative equal to zero to find the critical points:\[ \frac{2}{3} - \frac{2}{90}n = 0. \]Solve for \( n \):\[ \frac{2}{3} = \frac{2}{90}n \]\[ n = \frac{2}{3} \cdot \frac{90}{2} = 30. \]

Step 4: Test Critical Point for Maximum

To ensure that this point is a maximum, examine the second derivative:\[ E''(n) = \frac{d}{dn} \left( \frac{2}{3} - \frac{2}{90}n \right) = -\frac{2}{90}. \]Since \( E''(n) = -\frac{2}{90} \) is negative, the critical point at \( n = 30 \) is a maximum.

Key Concepts

Derivative Calculation

Calculating the derivative is an essential skill in finding the points where a function reaches its maximum or minimum values. In this case, the effectiveness function is given by \[ E(n) = \frac{2}{3}n - \frac{1}{90}n^2. \]Finding its derivative helps us determine where any changes in the effectiveness, represented by the function, begin to occur based on the number of times a viewer watches the commercial.We start by taking the first derivative of the function with respect to \( n \), which gives us:\[ E'(n) = \frac{d}{dn} \left( \frac{2}{3}n - \frac{1}{90}n^2 \right) = \frac{2}{3} - \frac{2}{90}n. \]Here are some key points to remember:

• The first derivative, \( E'(n) \), represents the rate of change of the effectiveness.
• Setting it to zero can help find the critical points, where these changes might peak or decline.

By calculating this first derivative, we pave the way to understanding how the number of viewings affects effectiveness.

Second Derivative Test

The second derivative test is a crucial step for confirming whether a function’s critical point is a maximum or minimum. This test utilizes the second derivative of the function, denoted as \( E''(n) \), to evaluate the concavity at the critical point. In this problem, after obtaining the critical point, we calculate:\[ E''(n) = \frac{d}{dn} \left( \frac{2}{3} - \frac{2}{90}n \right) = -\frac{2}{90}. \]Key facts about the second derivative test include:

• A positive second derivative indicates that the function is concave up, suggesting a minimum value at that point.
• A negative second derivative implies that the function is concave down, confirming a maximum at the critical point.

With \( E''(n) = -\frac{2}{90} \), a negative value confirms that when \( n = 30 \), the effectiveness is at its maximum, showcasing how often a viewer should watch the commercial for optimal impact.

Critical Points

Critical points are values of \( n \) where the first derivative of a function is zero or undefined. These points are the potential candidates for locations where a function might have a local maximum or minimum. In the given exercise, we set the first derivative \( E'(n) \) equal to zero:\[ \frac{2}{3} - \frac{2}{90}n = 0 \]Solving this equation provides the critical point:\[ n = 30 \]Key ideas about critical points:

• They can indicate maximum, minimum, or saddle points on the graph of the function.
• Further testing, such as through the second derivative test, is required to classify these points fully.

Thus, determining these points is essential in identifying where the effectiveness of a television commercial is optimized based on the number of times it is viewed.

Quadratic Functions

Quadratic functions are a type of polynomial function where the highest power of the variable is squared (raised to the power of two). They follow the general form \( ax^2 + bx + c \), representing parabolas when graphed. In this scenario, the effectiveness function\[ E(n) = \frac{2}{3}n - \frac{1}{90}n^2 \]is a quadratic function. Here are some characteristics of quadratic functions that are useful:

• They can open upward or downward, depending on the sign of the coefficient \( a \). In our case, since the coefficient of \( n^2 \) is negative, the parabola opens downward, indicating a maximum can occur.
• The vertex of the parabola represents the maximum or minimum point of the function, depending on its orientation.
• Completing the square or using derivatives can help locate the vertex or maximum/minimum points.

By understanding the nature of quadratic functions, identifying the effectiveness peak becomes more intuitive. In this problem, recognizing the downward opening of the parabola helps us confirm that the effectiveness reaches its peak at the vertex, which is the critical point deduced earlier.