Problem 334 - Calculus Volume 1

Question

Consider a pizzeria that sell pizzas for a revenue of \(R(x)=a x\) and costs Assume that \(\quad R(x)=15 x, \quad\) and \(C(x)=60+3 x+\frac{1}{2} x^{2}\). How many pizzas sold maximizes the profit?

Step-by-Step Solution

Answer

The profit is maximized when 12 pizzas are sold.

Step 1: Define the Profit Function

The profit function is given by the difference between the revenue function and the cost function. So, we define profit as \( P(x) = R(x) - C(x) \). Substituting the given functions, we have: \( P(x) = 15x - \left( 60 + 3x + \frac{1}{2}x^2 \right) \). Simplifying, \( P(x) = 15x - 60 - 3x - \frac{1}{2}x^2 \). Therefore, \( P(x) = -\frac{1}{2}x^2 + 12x - 60 \).

Step 2: Find the Critical Points

To find where the profit is maximized, take the derivative of the profit function with respect to \(x\) and set it to zero. So, we compute \( P'(x) = -x + 12 \). Setting \( P'(x) = 0 \) gives \( -x + 12 = 0 \), thus \( x = 12 \).

Step 3: Check the Second Derivative

To confirm that we have found a maximum (and not a minimum or inflection point), check the second derivative. The second derivative is \( P''(x) = -1 \). Since \( P''(x) = -1 < 0 \), the function is concave down at \( x = 12 \), indicating a maximum.

Key Concepts

Revenue Function

In the context of the given exercise, the revenue function represents the total income earned from selling a certain number of pizzas. Here, revenue is described by the function \( R(x) = 15x \). This means that each pizza sold generates $15 in revenue. The equation is linear with respect to the number of pizzas sold \( x \).
The revenue function is one of the foundational components when assessing business performance. It allows us to understand how changes in sales can impact overall earnings.

Cost Function

The cost function reflects the expenses associated with producing and selling pizzas. Given by \( C(x) = 60 + 3x + \frac{1}{2}x^2 \), it includes:

• Fixed costs, which are constant and equal to \(60. These costs do not change regardless of the number of pizzas sold.
• Variable costs, which depend on \( x \) and amount to \)3 per pizza. These increase linearly with each additional pizza produced.
• Additional costs represented by \( \frac{1}{2}x^2 \), indicating that costs grow at an increasing rate as production scales up.

Understanding the cost function is crucial as it helps determine the pricing strategy and assess the profitability of selling additional units.

Critical Points

Critical points are values of \( x \) where a function's derivative is zero or undefined. This is crucial for finding where the profit \( P(x) = R(x) - C(x) \) can potentially reach its maximum.
In the problem, the derivative of the profit function, \( P'(x) = -x + 12 \), is set to zero to solve for \( x \). This calculation helps identify potential points for maximum or minimum profit. Finding these points allows businesses to optimize operations, ensuring resources are not wasted and maximum profit is obtainable.

Second Derivative Test

The second derivative test helps determine the nature of critical points—whether they represent maxima, minima, or inflection points. After calculating the first derivative, we use \( P''(x) \) to assess the concavity of the profit function.
In the exercise, the second derivative \( P''(x) = -1 \) indicates that the function is always concave down \((P''(x) < 0)\). When \( x = 12 \), this negativity confirms a local maximum for the profit function. This step ensures confidence in decision-making regarding optimal production levels.