Problem 95 - Calculus and its applications

Question

A model for consumers' response to advertising is given by $$N(a)=2000+500 \ln a, \quad a \geq 1$$. where \(N(a)\) is the number of units sold and \(a\) is the amount spent on advertising, in thousands of dollars. a) How many units were sold after spending \(\$ 1000\) on advertising? b) Find \(N^{\prime}(a)\) and \(N^{\prime}(10)\). c) Find the maximum and minimum values, if they exist. d) Find \(\lim _{a \rightarrow \infty} N^{\prime}(a)\). Discuss whether it makes sense to continue to spend more and more on advertising.

Step-by-Step Solution

Answer

a) 2000 units. b) \(N'(a) = \frac{500}{a}\), \(N'(10) = 50\). c) Min value is 2000 at \(a=1\). No max value. d) \(\lim_{a \to \infty} N'(a) = 0\). Continued spending has diminishing returns.

Step 1: Find N(a) with a=1000

We know that the amount spent on advertising is given in thousands of dollars, so we have to substitute \( a = 1000 \) (i.e., \( a = 1 \)) into the function \( N(a) = 2000 + 500 \ln a \).First, convert \( 1000 \) dollars to thousands, which is \( a = 1 \). Substitute \( a = 1 \) into the function:\[ N(1) = 2000 + 500 \ln 1 \]Since \( \ln 1 = 0 \), we have:\[ N(1) = 2000 + 500 \times 0 = 2000 \]This means 2000 units were sold after spending \( \$1000 \) on advertising.

Step 2: Find the derivative N'(a)

To find \( N^{\prime}(a) \), we need to find the derivative of the function \( N(a) = 2000 + 500 \ln a \). The derivative of a constant is zero, and the derivative of \( \ln a \) is \( \frac{1}{a} \).So, \[ N^{\prime}(a) = 500 \cdot \frac{1}{a} = \frac{500}{a} \].

Step 3: Evaluate N'(10)

After finding the derivative, substitute \( a = 10 \) into the derivative:\[ N^{\prime}(10) = \frac{500}{10} = 50 \]This result tells us that the rate of change, or the increase in units sold with respect to each additional thousand dollars spent on advertising, is 50 units when \( \$10,000 \) is spent.

Step 4: Determine Maximum and Minimum Values

Look to assess whether the function \( N(a) \) has maximum or minimum values. In this function, \( N(a) = 2000 + 500 \ln a \) where \( a \geq 1 \). The \( \ln a \) function is continuous and increasing, so \( N(a) \) does not have a maximum.Since \( N(a) \) increases without bound with increasing \( a \), there cannot be a maximum. Its minimum when \( a \geq 1 \) is at \( a = 1 \), and \( N(1) = 2000 \).

Step 5: Evaluate Limit of N'(a) as a Approaches Infinity

To solve for \( \lim _{a \rightarrow \infty} N^{\prime}(a) \), observe the expression for \( N^{\prime}(a) = \frac{500}{a} \).As \( a \) approaches infinity, the expression \( \frac{500}{a} \) approaches 0.Thus, \[ \lim _{a \rightarrow \infty} N^{\prime}(a) = 0 \].This indicates that the additional effect or gain from advertising diminishes as more money is spent, suggesting that continual spending may not effectively increase sales significantly after a certain point.

Key Concepts

Derivative

In calculus, understanding derivatives is crucial as they reveal how a function changes at any given point. A derivative represents the rate at which one quantity changes concerning another. For the function in the problem, the number of units sold as a function of advertising spend is expressed as \(N(a) = 2000 + 500 \ln a\). Here, \(a\) is the amount spent on advertising in thousands of dollars. The derivative process involves finding \(N'(a)\), the rate of change of sales with respect to advertising spend.

• The derivative of a constant is zero.
• The derivative of \(\ln a\) is \(\frac{1}{a}\).

Therefore, the derivative of this function is \(N'(a) = \frac{500}{a}\). This derivative tells us how much the number of units sold increases for a slight increase in advertising spend at any specific point \(a\). In our specific application, when the advertising spend is increased from \(9,000 to \)10,000, we see a change of 50 units with the given function.

Limits

The concept of limits helps us understand the behavior of functions as they approach a particular point or infinity. In context, we look at the limit of the derivative \(N'(a)\), as \(a\) approaches infinity. With our derivative expression \(N'(a) = \frac{500}{a}\), calculating the limit gives an indication of what happens to the rate of sales as more money is spent on advertising.

• As \(a\) increases, \(\frac{500}{a}\) approaches 0.
• This signifies that the additional number of units sold diminishes with increased advertising.

Thus, \(\lim _{a \rightarrow \infty} N'(a) = 0\). As \(a\) grows larger, the impact on sales (in terms of units) becomes less significant, signaling a point where additional spending yields minimal return.

Consumer Response to Advertising

Consumer response to advertising often follows a model expressing how sales increase with more advertising. The function \(N(a) = 2000 + 500 \ln a\) demonstrates this, where buying responses are captured via a logarithmic relationship.

• Advertising impact is initially positive, increasing sales with more spend.
• A logarithmic model shows diminishing returns; initial increases are substantial but decrease over time.

This kind of model predicts that while initial gains from increased advertising are significant, eventually, further investment results in smaller returns. This assists businesses in deciding how much to spend on marketing efficiently, ensuring they do not surpass breakeven where expenditures outweigh benefits.

Optimization in Calculus

Optimization in calculus involves finding maximum or minimum values of functions to make decisions that yield the best results, such as maximizing profit or minimizing cost. The function \(N(a)\) reflects unit sales tied to spending, allowing us to explore if there's an optimal advertising spend.

• Examine for maxima/minima by analyzing the derivative \(N'(a)\).
• The function \(N(a) = 2000 + 500 \ln a\) continuously increases, lacking a maximum under the condition \(a \geq 1\).
• A minimum exists at \(a = 1\), yielding \(N(1) = 2000\) units.

By understanding when additional spending becomes inefficient, businesses can prioritize budgets more effectively, focusing on strategies providing greater yield rather than pursuing diminishing returns.