Problem 55 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

SALES FROM ADVERTISING It is estimated that if \(x\) thousand dollars are spent on advertising. approximately \(Q(x)=50-40 e^{-0.1 x}\) thousand units of a certain commodity will be sold. a. Sketch the sales curve for \(x \geq 0\). b. How many units will be sold if no money is spent on advertising? c. How many units will be sold if \(\$ 8,000\) is spent on advertising? d. How much should be spent on advertising to generate sales of 35,000 units? e. According to this model, what is the most optimistic sales projection?

Step-by-Step Solution

Answer

a) Sketch the graph. b) 10,000 units. c) 32,000 units. d) ~$9,803 e) 50,000 units.

Step 1 - Sketching the Sales Curve

First, consider the function for the sales curve: \[ Q(x) = 50 - 40 e^{-0.1x} \] To sketch the curve for \( x \geq 0 \), plot several points by substituting different values of \( x \). Note that as \( x \) increases, the term \( e^{-0.1x} \) decreases towards 0.

Step 2 - Units Sold with No Advertising

Substitute \( x = 0 \) into the sales function: \[ Q(0) = 50 - 40 e^{-0.1 \times 0} = 50 - 40 \times 1 = 10 \] Thus, if no money is spent on advertising, 10,000 units will be sold.

Step 3 - Units Sold with \$8,000 Advertising

Substitute \( x = 8 \) into the sales function: \[ Q(8) = 50 - 40 e^{-0.1 \times 8} = 50 - 40 e^{-0.8} \] Calculate \( e^{-0.8} \approx 0.4493 \), then: \[ Q(8) = 50 - 40 \times 0.4493 \approx 50 - 18 = 32 \] Therefore, if \$8,000 is spent on advertising, 32,000 units will be sold.

Step 4 - Advertising to Generate 35,000 Units

Set \( Q(x) = 35 \) and solve for \( x \): \[ 35 = 50 - 40 e^{-0.1 x} \] Rearrange to solve for the exponential term: \[ 40 e^{-0.1 x} = 15 \] \[ e^{-0.1 x} = \frac{15}{40} = 0.375 \] Take the natural logarithm of both sides: \[ -0.1 x = \text{ln}(0.375) \approx -0.9803 \] Solve for \( x \): \[ x = \frac{-0.9803}{-0.1} = 9.803 \] Approximately \$9,803 should be spent on advertising to generate sales of 35,000 units.

Step 5 - Most Optimistic Sales Projection

As \( x \to \infty \), \( e^{-0.1x} \) approaches 0. Thus, the maximum sales would approach: \[ Q(x) = 50 - 40 \times 0 = 50 \] The most optimistic sales projection is 50,000 units.

Key Concepts

exponential functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In this problem, the sales function is given as: \[ Q(x) = 50 - 40 e^{-0.1x} \]Here, the term involving the exponential function is \( e^{-0.1x} \). As we increase x, the exponential term \( e^{-0.1x} \) decreases towards 0. This is an important property of exponential decay: it diminishes quickly at first and then levels off. Understanding this helps us realize that increases in advertising spending have diminishing returns on sales.

sales projection

Sales projection involves predicting future sales based on certain factors, such as advertising. In this exercise, the sales function tells us how many units will be sold based on dollars spent on advertising. By substituting different values of \( x \) (representing thousands of dollars spent), we can see how sales numbers change. For example:

• When \( x = 0 \): No money spent on advertising implies sales of 10,000 units.
• When \( x = 8 \): Spending \$8,000 on advertising leads to sales of approximately 32,000 units.

Understanding these projections helps businesses allocate their advertising budgets wisely to maximize sales.

advertising impact

Advertising impact measures how effective advertising spending is at increasing sales. In our equation,\( Q(x) = 50 - 40 e^{-0.1x} \), the impact of advertising is seen through the change in sales as we increase \( x \). We can observe several key points:

• Initial impact: At lower levels of spending, each additional dollar has a significant effect. For instance, increasing from \$0 to thousands of dollars quickly boosts sales.
• Diminishing returns: As spending increases, the rate of sales increase slows down. Spending more continues to boost sales, but the gains are smaller.
• Maximum potential: Eventually, as spending approaches extremely high values, sales near its maximum limit (50,000 units in this case).

This understanding helps businesses optimize their advertising spend to achieve the best possible outcomes.

unit sales calculation

To calculate unit sales, we use the given sales function. Let's break down the steps:

• First, identify the value of \( x \) representing the thousands of dollars spent on advertising.
• Substitute this value into the function \( Q(x) = 50 - 40 e^{-0.1x} \).
• Simplify the expression to get the sales in thousands of units.

For example, to find the sales with \$8,000 spent on advertising:

• Substituting \( x = 8 \) into \( Q(x) \), we get: \( Q(8) = 50 - 40 e^{-0.8} \).
• Calculate \( e^{-0.8} \approx 0.4493 \).
• Finally, \( Q(8) = 50 - 40 \times 0.4493 \approx 32 \), meaning 32,000 units.

By understanding how to substitute and calculate these values, you can predict sales effectively based on advertising spending.