Problem 43 - College Algebra and Calculus: An Applied Approach

Question

A fast-food restaurant determines the cost and revenue models for its hamburgers. \(C=0.6 x+7500, \quad 0 \leq x \leq 50,000\) \(R=\frac{1}{20,000}\left(65,000 x-x^{2}\right), \quad 0 \leq x \leq 50,000\) (a) Write the profit function for this situation. (b) Determine the intervals on which the profit function is increasing and decreasing. (c) Determine how many hamburgers the restaurant needs to sell to obtain a maximum profit. Explain your reasoning.

Step-by-Step Solution

Answer

The profit function is \( P(x) = \frac{x^2}{20,000} +3.25x - 19500 \). The profit function is increasing for \( x < 32,500 \) and decreasing for \( x > 32,500 \). The restaurant needs to sell 32,500 hamburgers to achieve maximum profit.

Step 1: Determine the profit function

The profit function is the difference between the revenue and the cost. Therefore, using the given cost \(C\) and revenue function \(R\) to find the profit function as: \( P(x) = R(x) - C(x) = \frac{1}{20,000}(65,000x - x^2) - (0.6x + 7500)\)

Step 2: Simplify the profit function

Simplify the equation as: \( P(x) = \frac{1}{20,000}(65,000x - x^2) - 0.6x - 7500 \). Note that in order to subtract 0.6x and 7500, they need to be multiplied by 20,000 first to the same term on the left (because of the division by 20,000). Hence, \( P(x) = 3.25x - \frac{x^2}{20,000} - 12,000 - 7500 \) which simplifies to \( P(x) = \frac{x^2}{20,000} +3.25x - 19500 \)

Step 3: Determine where the profit function is increasing or decreasing

To find where the profit function is increasing or decreasing, the first derivative of the profit function should be taken and set to zero. This gives possible points of maximum or minimum (turning points). Hence, the first derivative is: \( P'(x) = -\frac{2x}{20000} + 3.25 \). Setting \( P'(x) = 0 \), we get \( x = 32,500 \)

Step 4: Determine the maximum profit

To know whether this point is a maximum or minimum, the second derivative test can be used - if the second derivative at this point is negative, it's a maximum, otherwise it's a minimum. The second derivative of the profit function is: \( P''(x) = -\frac{2}{20000} \) which is always negative since it's a constant. Hence, the point \( x = 32,500 \) is a maximum.

Step 5: Contextual interpretation

From the results, to obtain the maximum profit, the fast-food restaurant needs to sell 32,500 hamburgers. This is the point where the profit changes from increasing to decreasing, making it the point of maximum profit.

Key Concepts

Revenue Model

In the context of economics, a revenue model defines how much income a business can generate by selling a product. Here, the restaurant's revenue model is given by the function: \[ R(x) = \frac{1}{20,000}(65,000x - x^2) \]. This function shows that revenue depends on the number of hamburgers sold, denoted as \( x \).
The formula includes:

• Linear term: \( 65,000x \) - showing direct proportionality to the number sold.
• Quadratic term: \( -x^2 \) - reflecting diminishing returns or the potential decrease in revenue after a certain number of hamburgers are sold.

The division by 20,000 adjusts these quantities into an appropriate model for revenue. This model suggests that initially, as sales increase, revenue increases, but beyond a certain point, adding more sales might decrease total revenue, perhaps due to pricing, supply limitations, or market saturation.

Cost Model

The cost model explains how much it costs the restaurant to produce hamburgers. It’s described by the function:\[ C(x) = 0.6x + 7500 \]. This function states that:

• Variable cost per hamburger is \( 0.6 \), representing cost that changes with each additional hamburger sold.
• Fixed costs, represented by 7500, are constant expenses that do not vary with production volume, such as rent or salaries.

Understanding cost is crucial for calculating the profit function. Costs must be balanced with revenue to ensure a business remains profitable. By identifying how costs rise with hamburger production, the restaurant can make more informed pricing and production decisions.

First Derivative

Derivatives are a fundamental tool in calculus used to find rates of change. For this exercise, the first derivative of the profit function helps determine where the profit is increasing or decreasing. The first derivative of the profit function \( P(x) \) is:\[ P'(x) = -\frac{2x}{20000} + 3.25 \].
Setting \( P'(x) = 0 \) finds the critical points. Solving gives \( x = 32,500 \).
This result tells us that the profit changes rate at 32,500 hamburgers sold.

• If \( P'(x) > 0 \), profit is increasing, potentially maximizing if it changes from positive to negative.
• If \( P'(x) < 0 \), profit is decreasing.

These critical points are pivotal in determining the optimal level of production for maximum profit.

Second Derivative

The second derivative provides insight into the concavity of a function and helps identify whether a critical point is a maximum or minimum. Calculating the second derivative of the profit function gives:\[ P''(x) = -\frac{2}{20000} \].This shows a constant negative value.
In economic terms, a negative second derivative suggests the function is concave downwards, indicating a maximum point.
In this problem, since \( P''(x) < 0 \) for all \( x \), it confirms that \( x = 32,500 \) is indeed a maximum point for the profit function. Thus, selling 32,500 hamburgers will yield the highest profit for the restaurant.