Problem 12 - 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 \(p(100)=\$ 20.09\) . a. Find \(p(x)\) by solving the following initial value problem: $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d p}{d x}=-\frac{1}{100} p} \\ {\text { Initial condition: }} & {p(100)=20.09}\end{array} $$ b. Find the unit price \(p(10)\) for a 10 -unit order and the unit price \(p(90)\) for a 90 -unit order. c. The sales department has asked you to find out if if is discounting so much that the firm's revenue, \(r(x)=x \cdot p(x)\) will actually be less for a 100 -unit order than, say, for a \(90-\) unit order. Reassure them by showing that \(r\) has its maximum value at \(x=100\) . d. Graph the revenue function \(r(x)=x p(x)\) for \(0 \leq x \leq 200\) .

Step-by-Step Solution

Answer

The revenue function's maximum is at \( x = 100 \).

Step 1: Identify the Differential Equation

We are given the differential equation \( \frac{d p}{d x} = -\frac{1}{100} p(x) \) with the initial condition \( p(100) = 20.09 \). This is a first-order linear differential equation that needs to be solved.

Step 2: Solve the Differential Equation

The general form for a first-order linear differential equation is solved by separation of variables. We rearrange the expression:\[ \frac{d p}{p} = -\frac{1}{100} \, dx \]Integrating both sides, we get:\[ \ln |p| = -\frac{1}{100}x + C \]Exponentiating both sides, we find:\[ p(x) = e^{C}e^{-\frac{x}{100}} \]

Step 3: Apply the Initial Condition

We apply the initial condition \( p(100) = 20.09 \) to find \( C \):\[ p(100) = e^{C}e^{-1} = 20.09 \]Thus, \( e^{C} = 20.09 \, e \).Therefore, the solution is:\[ p(x) = 20.09 \, e^{1 - \frac{x}{100}} \]

Step 4: Calculate \( p(10) \) and \( p(90) \)

Using the function derived, calculate \( p(10) \) and \( p(90) \):\[ p(10) = 20.09 \, e^{1 - \frac{10}{100}} = 20.09 \, e^{0.9} \approx 49.36 \, (check with a calculator) \]\[ p(90) = 20.09 \, e^{1 - \frac{90}{100}} = 20.09 \, e^{0.1} \approx 22.22 \, (check with a calculator) \]

Step 5: Calculate Revenue Function \( r(x) \)

We have \( r(x) = x \, p(x) \). Substitute for \( p(x) \):\[ r(x) = x \, (20.09 \, e^{1 - \frac{x}{100}}) \]The function \( r(x) = 20.09 \, x \, e^{1 - \frac{x}{100}} \) is our revenue function, which needs analysis to check its behavior.

Step 6: Analyze Revenue for Maximum Value at \( x = 100 \)

To find where \( r(x) \) attains maximum value, consider the derivative \( r'(x) \) and solve it:\[ r'(x) = \frac{d}{dx}(20.09 \, x \, e^{1 - \frac{x}{100}}) \approx 0 \]Differentiate using the product rule, solve for critical points to find that \( x = 100 \) gives a maximum after verifying by second derivative or graph.

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

Plot \( r(x) = 20.09 \, x \, e^{1 - \frac{x}{100}} \) for \( 0 \leq x \leq 200 \) using a graphing tool or software, marking the peak at \( x = 100 \) to visualize how revenue changes with respect to different unit orders.

Key Concepts

Revenue Function

The revenue function serves as a crucial aspect of business economics and calculus. In simple terms, a revenue function, denoted as \( r(x) \), is used to determine how much income or revenue a firm generates from selling a certain quantity of goods, where \( x \) represents the quantity sold.

• The revenue function can be mathematically expressed as \( r(x) = x \cdot p(x) \), where \( p(x) \) is the price function indicating how much each unit is sold for.
• The change in revenue depends on fluctuations in quantity \(x\) and unit price \(p(x)\).

Understanding the revenue function allows firms to predict sales outcomes and optimize pricing strategies.
By leveraging the revenue function, companies aim to find the balance point where revenue is maximized with respect to quantity sold.

Continuous Discounting

Continuous discounting is a concept where the price of a product is adjusted using a mathematical process that decreases the price based on the quantity ordered. It's particularly useful for encouraging customers to order larger quantities by offering better pricing per unit as orders increase.

• The decrease rate in our context was \( \$0.01 \) per unit ordered, presented as a differential equation \( \frac{d p}{d x} = -\frac{1}{100} p \).
• This mathematical representation helps model how each additional unit changes the price dynamically.

By incorporating continuous discounting, businesses can create a structured pricing model that entices larger purchases.
Such models are utilized to determine optimal discount rates that maximize sales while preventing excessive revenue loss from discounts.

Initial Value Problem

An initial value problem (IVP) is a differential equation accompanied by a given initial condition. It's a tool used to find specific solutions to differential equations that provide crucial information about a scenario,
such as predicting price changes in our continuous discounting scenario.

• In this situation, the initial condition is \( p(100) = 20.09 \), representing the unit price at 100 units ordered.
• The differential equation is solved by applying the initial condition to discover the function that represents \( p(x) \), the price per unit ordered \( x \).

Solving an IVP gives us an explicit formula to predict unit prices for any order amount, enhancing decision-making in pricing strategies.

Optimization in Calculus

Optimization in calculus refers to finding the maximum or minimum values of a function within a given domain. This concept is vital for businesses needing to identify optimal points for pricing, revenue, or other important variables.

• For our revenue function \( r(x) = 20.09 \cdot x \cdot e^{1 - \frac{x}{100}} \), optimization involves finding the quantity \(x\) that maximizes \(r(x)\).
• To achieve this, we calculate the derivative of \( r(x) \) and solve for the critical points that indicate maximum or minimum values.
• In our example, the maximum revenue occurs at \( x = 100 \), verified using calculus techniques such as the second derivative test.

Understanding optimization enables businesses to make informed decisions that drive profitability, such as setting the right order quantities that maximize revenue.