Problem 36 - Thomas Calculus

Question

Continuous price discounting To encourage buyers to place 100 -unit orders, your firm's sales department applies a continuous discount that makes the unit price a function \(p(x)\) of the number of units \(x\) ordered. The discount decreases the price at the rate of \(\$ 0.01\) per unit ordered. The price per unit for a 100 -unit order is \(p(100)=\$ 20.09 .\) \begin{equation} \begin{array}{c}{\text { a. Find } p(x) \text { by solving the following initial value problem: }} \\ {\text { Differential equation: } \quad \frac{d p}{d x}=-\frac{1}{100} p} \\ {\text { Initial condition: } \quad p(100)=20.09}\\\\{\text { b. Find the unit price } p(10) \text { for a } 10 \text { -unit order and the unit price }} \\ {p(90) \text { for a } 90 \text { -unit order. }}\\\\{\text { c. The sales department has asked you to find out if it is dis- }} \\ {\text { counting so much that the firm's revenue, } r(x)=x \cdot p(x), \text { will }} \\ {\text { actually be less for a } 100 \text { -unit order than, say, for a } 90 \text { -unit }}\\\\{\text { order. Reassure them by showing that } r \text { has its maximum }} \\ {\underline{\phantom{xxx}} }\\\\{\text { value at } x=100 \text { . }} \\ {\text { d. Graph the revenue function } r(x)=x p(x) \text { for } 0 \leq x \leq 200 \text { . }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The unit prices are \( p(10) \approx 49.07 \) and \( p(90) \approx 22.19 \). Maximum revenue occurs at \( x=100 \).

Step 1: Solve the Differential Equation

The differential equation given is \( \frac{dp}{dx} = -\frac{1}{100}p \). This is a separable differential equation. We separate the variables as \( \frac{1}{p}dp = -\frac{1}{100}dx \). Integrating both sides gives \( \ln|p| = -\frac{1}{100}x + C \), where \( C \) is the integration constant.

Step 2: Solve for the Integration Constant

From Step 1, we have \( \ln|p| = -\frac{1}{100}x + C \). Applying the initial condition \( p(100) = 20.09 \), we substitute \( x=100 \) to get \( \ln(20.09) = -1 + C \), which gives \( C = \ln(20.09) + 1 \). Thus, the solution becomes \( \ln|p| = -\frac{1}{100}x + \ln(20.09) + 1 \).

Step 3: Express \( p(x) \) Explicitly

Exponentiating both sides of the equation from Step 2, we get \( p(x) = e^{-\frac{1}{100}x} \cdot e^{\ln(20.09) + 1} = 20.09e \cdot e^{-\frac{1}{100}x} \). Simplifying, we have \( p(x) = 20.09e^{-\frac{1}{100}(x-100)} \).

Step 4: Calculate \( p(10) \)

Substitute \( x=10 \) into the equation from Step 3: \( p(10) = 20.09e^{-\frac{1}{100}(10-100)} = 20.09e^{0.9} \). Compute this value to find \( p(10) \approx 49.07 \).

Step 5: Calculate \( p(90) \)

Substitute \( x=90 \) into the equation from Step 3: \( p(90) = 20.09e^{-\frac{1}{100}(90-100)} = 20.09e^{0.1} \). Compute this value to find \( p(90) \approx 22.19 \).

Step 6: Analyze Revenue \( r(x) \)

The revenue function is \( r(x) = x \cdot p(x) = x \cdot 20.09e^{-\frac{1}{100}(x-100)} \). Differentiate to find critical points: \( r'(x) = 20.09(e^{-\frac{1}{100}(x-100)} - \frac{x}{100}e^{-\frac{1}{100}(x-100)}) \). Set \( r'(x) = 0 \) and solve for \( x \) to find \( x = 100 \), which confirms the maximum.

Step 7: Graph \( r(x) \)

To graph \( r(x) = x \cdot 20.09e^{-\frac{1}{100}(x-100)} \) for \( 0 \leq x \leq 200 \), use a graphing tool. Note that the graph shows a peak at \( x=100 \), confirming maximum revenue at this point.

Key Concepts

Continuous Discounting

Continuous discounting is a method used to offer price reductions based on the quantity of goods ordered. Here, the unit price, denoted as a function of the number of units ordered (\(p(x)\)), decreases at a continuous rate. In the given exercise, the reduction rate is \(-\\(0.01\) per unit ordered.
This implies that as more units are ordered, the price per unit is discounted consistently. Such a model is helpful for companies to incentivize bulk purchases by reducing the cost per item. The initial condition given is \(p(100)=\\)20.09\), which means that at 100 units, each unit costs \$20.09.
This concept helps in determining how pricing can be adjusted for different order sizes to achieve specific financial objectives like maximizing revenue.

Revenue Maximization

Revenue maximization is about finding the price and quantity combination that generates the highest possible revenue for a business. In this context, revenue (\(r(x)\)) is defined as the product of the number of units sold \(x\) and the price per unit \(p(x)\).
Calculating revenue involves multiplying these two factors: \(r(x) = x \cdot p(x)\).
The task is to determine if the firm's revenue with a 100-unit order is higher than that with a 90-unit order.
To find the maximum revenue point, we differentiate the revenue function with respect to \(x\) and set the derivative \(r'(x)\) to zero to find critical points. Through analysis, it is determined that \(x = 100\) is indeed where maximum revenue occurs, indicating successful revenue maximization at this point.
This demonstrates the effectiveness of continuous discounting in achieving revenue optimization.

Separable Differential Equations

Separable differential equations are a class of differential equations where variables can be separated onto different sides of the equation, allowing for straightforward integration.
In the exercise, we encountered the differential equation \(\frac{dp}{dx} = -\frac{1}{100}p\). This is separable because it can be rearranged to \(\frac{1}{p}dp = -\frac{1}{100}dx\), separating the variables \(p\) and \(x\).
Integrating both sides results in a solution that relates \(p\) to \(x\), in this case giving us \(\ln|p| = -\frac{1}{100}x + C\), where \(C\) is an integration constant determined using an initial condition.
Solving this kind of equation forms the basis for many applications in physical and economic settings, such as continuous discounting, where the price changes continuously with the ordered quantity.