Problem 31 - Calculus and its applications

Question

Suppose that SpryBorg Inc. introduces a new computer game in Houston using television advertisements. Surveys show that \(P \%\) of the target audience buys the game after \(x\) ads are broadcast, satisfying $$ P(x)=\frac{100}{1+49 e^{-0.13 x}} $$ a) What percentage buys the game without seeing a TV \(\operatorname{ad}(x=0) ?\) b) What percentage buys the game after the ad is run 5 times? 10 times? 20 times? 30 times? 50 times? 60 times? c) Find the rate of change, \(P^{\prime}(x)\). d) Sketch a graph of the function.

Step-by-Step Solution

Answer

2% buy without ads; calculate with formula for others. Derive \( P'(x) \) for rate of change.

Step 1: Determine the initial percentage (x=0)

To find the percentage of people who buy the game without seeing any TV ads, substitute x = 0 into the given function. So we calculate \( P(0) = \frac{100}{1 + 49 e^{-0.13 \cdot 0}} \). Since \( e^0 = 1 \), the expression simplifies to \( P(0) = \frac{100}{1 + 49} = \frac{100}{50} = 2 \). Thus, 2% of the target audience buys the game without seeing any TV ads.

Step 2: Calculate percentage for specific numbers of ads

Substitute \( x = 5, 10, 20, 30, 50, 60 \) into the expression for \( P(x) \) to find the percentage for those specific numbers of ads. - For \( x = 5 \), \( P(5) = \frac{100}{1 + 49 e^{-0.13 \cdot 5}} \).- For \( x = 10 \), \( P(10) = \frac{100}{1 + 49 e^{-0.13 \cdot 10}} \).- Continue this for \( x = 20, 30, 50, 60 \), calculating each time using a calculator.

Step 3: Derive the rate of change, P'(x)

To find the rate of change \( P'(x) \), differentiate \( P(x) = \frac{100}{1 + 49 e^{-0.13 x}} \) using the quotient rule: \[ P'(x) = \frac{d}{dx} \left( \frac{100}{1 + 49 e^{-0.13 x}} \right) \]The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = 100 \) and \( v = 1 + 49 e^{-0.13x} \). Calculate \( u' = 0 \) and \( v' = -49 \times 0.13 \times e^{-0.13 x} \), and put these into the quotient rule formula.

Step 4: Sketch the graph of the function

To sketch the graph of \( P(x) \), note that it is a logistic function with an asymptote at 100% as x increases. Based on calculations of \( P(x) \), determine key points such as at \( x = 0, 5, 10, \text{etc.} \) and plot these. The graph starts at 2% for \( x = 0 \), increases rapidly for low values of \( x \), and levels off near 100% as \( x \) increases.

Key Concepts

Logistic Function

The logistic function is an important concept often used in various fields, such as biology, economics, and here, in marketing. It is particularly useful to model scenarios where growth is initially exponential but eventually slows down as it approaches a maximum limit. In our case, the function is defined as:

• \[ P(x)=\frac{100}{1+49 e^{-0.13 x}} \]

This equation shows how the percentage \(P\) of a target audience buying a new computer game changes with the number of advertisements \(x\). Let's break it down:

• The numerator represents the maximum limit of 100%, which is the situation where every potential customer buys the game.
• The denominator consists of a term that decreases exponentially, modified by the constant \(49\), and \(e^{-0.13x}\), which accounts for how each ad's effectiveness diminishes as you saturate the audience.

The logistic function beautifully depicts these real-world constraints, depicting a rapid initial increase in the percentage, which slows as it inches closer to its upper limit of 100%.

Differentiation

Differentiation is a key concept in calculus that helps us find the rate at which a quantity changes. For the logistic function describing SpryBorg Inc.'s ad effectiveness, differentiation allows us to understand how quickly the purchase percentage changes with each additional advertisement.To differentiate the logistic function \(P(x)=\frac{100}{1+49 e^{-0.13 x}}\), we use the quotient rule:

• The quotient rule formula is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
• In this context, \(u = 100\) and \(v = 1 + 49e^{-0.13x}\).
• \(u'\) is the derivative of \(u\), which is zero because \(100\) is a constant.
• \(v'\) is the derivative of \(v\), calculated as \(-49 \times 0.13 \times e^{-0.13 x}\).

By substituting these components into the quotient rule, we can compute \(P'(x)\), highlighting how the ad's impact diminishes over time with each additional airing.

Rate of Change

The rate of change is pivotal when analyzing how a function responds to changes in its variables. In this scenario, the rate of change helps us understand how the percentage of the target audience buying the game shifts as more TV ads are broadcasted.For our logistic model, calculating the derivative \(P'(x)\) provides insights as to how \(P\%\) of the audience changes as \(x\), the number of ads, increases. The derivative reveals key information:

• When \(P'(x) > 0\), the percentage of buyers is increasing.
• As \(x\) grows larger, \(P'(x)\) tends to decrease, indicating a tapering effect where additional ads bring fewer new buyers.
• This deceleration is typical in logistic growth, illustrating diminishing returns as saturation is reached.

The logistic function, paired with its derivative, deepens our understanding of strategic marketing planning, showing how additional ads have progressively less impact on customer purchases.

Graph Sketching

Graphing the logistic function \(P(x)\) helps visualize this entire process, illustrating the relationship between the number of ads broadcast and the percentage of the audience that purchases the game. Here are the key features of the graph:

• The graph starts at 2% for \(x = 0\), the point at which no ads have been shown. This indicates that even without advertisements, a small segment of the market adopts the game due to other factors like word of mouth.
• As \(x\) increases, the graph rises sharply, depicting an accelerated growth as more people are exposed to the ads and buy the game.
• However, the growth isn't indefinite. It begins to plateau, approaching an asymptote near 100%, indicating the saturation point where most of the target audience has already bought the game.
• This flattening phase showcases the diminishing returns typical in population growth models, where further increases in \(x\) yield smaller increases in \(P(x)\).

Visually interpreting the logistic function through graph sketching enhances comprehension, highlighting transition points and saturation throughout the marketing campaign.