Problem 333 - Calculus Volume 1

Question

Consider a pizzeria that sell pizzas for a revenue of \(R(x)=a x\) and costs \(C(x)=b+c x+d x^{2},\) where \(x\) represents the number of pizzas. Assume that \(R(x)=10 x\) and \(C(x)=2 x+x^{2}\). How many pizzas sold maximizes the profit?

Step-by-Step Solution

Answer

The profit is maximized when 4 pizzas are sold.

Step 1: Define the Profit Function

The profit function is defined as the difference between revenue and costs: \( P(x) = R(x) - C(x) \). Substitute the given functions: \( R(x) = 10x \) and \( C(x) = 2x + x^2 \) into the equation to get the profit function: \( P(x) = 10x - (2x + x^2) \). Simplify the expression to \( P(x) = 8x - x^2 \).

Step 2: Set Derivative to Zero

To find the number of pizzas that maximizes profit, take the derivative of the profit function \( P(x) \) and set it equal to zero. The derivative of \( P(x) = 8x - x^2 \) is \( P'(x) = 8 - 2x \). Set the derivative equal to zero: \( 8 - 2x = 0 \).

Step 3: Solve for x

Solve the equation \( 8 - 2x = 0 \) for \( x \). Add \( 2x \) to both sides to get \( 8 = 2x \), then divide both sides by 2 to find \( x = 4 \).

Step 4: Verify Maximum Condition

To verify that \( x = 4 \) is a maximum, check the second derivative \( P''(x) \). The second derivative is \( P''(x) = -2 \), which is less than zero, indicating that the function is concave down and \( x = 4 \) is a maximum.

Key Concepts

Revenue Function

The concept of a revenue function is fundamental in understanding profit maximization. Simply put, the revenue function, denoted as \( R(x) \), expresses the total revenue earned by selling \( x \) units of a product. In our pizzeria example, pizzas sell at a price leading to a revenue function \( R(x) = 10x \). This equation implies that for every additional pizza sold, the revenue increases by 10 monetary units.
Hence, the coefficient in the function represents the unit price of the product. Understanding the revenue function is the first step towards identifying how various sales levels impact total revenue.

Cost Function

The cost function, denoted as \( C(x) \), represents the total cost associated with producing \( x \) units. In the scenario of our pizzeria, the cost function is expressed as \( C(x)=2x+x^2 \).
This function consists of both linear and quadratic terms. The linear term, 2x, might represent variable costs that change with every pizza made, while the quadratic term xpresents non-linear costs often related to increasing inefficiencies at higher production levels.
Understanding the components of the cost function is crucial. They indicate how total costs may fluctuate as production quantities change.

Derivative

Derivatives are a core mathematical concept used to determine the rate of change. In the context of profit maximization, the derivative of the profit function helps find critical points where profit might be maximized.
For our pizzeria, we derive the function \( P(x) = 8x - x^2 \) to get \( P'(x) = 8 - 2x \). This derivative represents changes in profit as more pizzas are sold. Setting \( P'(x) = 0 \) determines the quantity where profit stops increasing and identifies potential maxima or minima. Thus, derivatives serve as a vital tool for analyzing and optimizing economic functions.

Second Derivative Test

The second derivative test is a mathematical method for confirming whether a critical point is a maximum or minimum. Once the first derivative is set to zero and solved for \( x \), the second derivative offers additional insights.
For our pizza problem, the second derivative \( P''(x) = -2 \), being less than zero, indicates the function is concave down. This concavity signals that the critical point calculated is indeed a maximum.
Understanding the second derivative test empowers you to not just find potential optima, but confirm the nature of these points in terms of maximizing, minimizing, or merely being stationary.