Problem 63 - Calculus and its applications

Question

Eloy Chutes determines that the marginal revenue, \(R^{\prime},\) in dollars per unit, from selling the \(x\) th parachute is given by \(R^{\prime}(x)=x^{2}-3\) a) Find the total-revenue function, \(R\), assuming that \(R(0)=0\) b) Why is \(R(0)=0\) a reasonable assumption?

Step-by-Step Solution

Answer

a) The total revenue function is \( R(x) = \frac{x^3}{3} - 3x \). b) \( R(0)=0 \) is reasonable as no sales mean no revenue.

Step 1: Integrate the Marginal Revenue Function

To find the total revenue function, we need to integrate the marginal revenue function, \( R'(x) = x^2 - 3 \). Integration of \( x^2 - 3 \) with respect to \( x \) yields: \[ R(x) = \int (x^2 - 3) \, dx = \frac{x^3}{3} - 3x + C \] where \( C \) is the constant of integration.

Step 2: Apply Initial Condition

We use the initial condition \( R(0) = 0 \) to determine the constant of integration \( C \). Substitute \( x = 0 \) into the total revenue function: \[ R(0) = \frac{0^3}{3} - 3 \times 0 + C = 0 \] Simplifying, we find \( C = 0 \).

Step 3: Write the Total Revenue Function

Substituting \( C = 0 \) back into the total revenue function, we get: \[ R(x) = \frac{x^3}{3} - 3x \] This is the total revenue function based on the given marginal revenue.

Step 4: Reasoning for Initial Condition

Since \( R(0) = 0 \), it implies that when no parachutes are sold, the total revenue is zero. It is reasonable because no products sold would naturally result in no revenue generated.

Key Concepts

Marginal Revenue

Marginal revenue is a critical concept in economics that helps businesses determine the additional income they receive from selling one more unit of a product. It provides insights into how revenue changes as sales volume varies. The formula for marginal revenue derived in this exercise is given by: \[ R'(x) = x^2 - 3 \] This equation represents how the extra revenue from selling the \(x\)th parachute is calculated. To gain a deeper understanding:

• The term \(x^2\) suggests that the revenue increase accelerates as more parachutes are sold, indicating scalability in revenue gain at higher sales levels.
• The constant \(-3\) indicates a fixed deduction or base expense from every sale, representing costs or diminishing returns.

Using marginal revenue functions like this enables companies to strategize effectively on pricing and production quantity. Making informed decisions can maximize profit and market potential.

Total Revenue Function

To gain a comprehensive picture of revenue across varying sales volumes, integrating the marginal revenue function gives us the total revenue function. Integration is a process in calculus for finding the accumulation effect of a rate of change, like how revenue accumulates for each additional sale. Given the marginal revenue function: \[ R'(x) = x^2 - 3 \] By integrating, we find the total revenue function: \[ R(x) = \int (x^2 - 3) \, dx = \frac{x^3}{3} - 3x + C \] Where:

• \( \frac{x^3}{3} \) represents the growing revenue component proportional to the cube of sales, highlighting how revenue scales exponentially.
• \(-3x\) represents the linear reduction due to consistent costs or diminishing returns per unit.
• \(C\) is an integration constant that adjusts the formula based on initial conditions.

Calculating the total revenue function helps businesses forecast total sales revenue and ensure efficient production and marketing strategies.

Initial Condition in Calculus

Calculus often uses initial conditions to determine constants of integration, which adjust general solutions to specific cases. Identifying and using initial conditions allows one to tailor the mathematical solutions for real-world problems accurately. In the given exercise, the initial condition is: \[ R(0) = 0 \] This means when zero parachutes are sold, the total revenue is \(0\). This is a straightforward concept rooted in the reality that no sales result in no revenue. The importance of this initial condition includes:

• Solving for the constant \(C\): It restricts the general solution to fit specific starting points or conditions.
• Ensuring logical consistency: The assumption mirrors real-world situations perfectly, making it reasonable and acceptable in calculations.

Understanding and applying initial conditions in calculus is essential to customizing solutions and enhancing precision in mathematical modeling and prediction scenarios.