Problem 36 - Thomas Calculus in SI Units

Question

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: $$\begin{aligned} &\text { Differential equation: } \quad \frac{d p}{d x}=-\frac{1}{100} p\\\ &\text { Initial condition: } \quad p(100)=20.09 \end{aligned}$$ 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 it 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 unit prices are: \( p(10) = 20.09e^{0.9} \), \( p(90) = 20.09e^{0.1} \), and revenue is maximized at \( x = 100 \).

Step 1: Solve the Differential Equation

We need to solve \( \frac{d p}{d x} = -\frac{1}{100} p \) with the given initial condition \( p(100) = 20.09 \). This is a separable differential equation. Separating the variables, we get \( \frac{dp}{p} = -\frac{1}{100} dx \).Integrating both sides, we have \( \ln |p| = -\frac{1}{100}x + C \). Exponentiating both sides, we get \( |p| = e^C e^{-\frac{1}{100}x} \) or equivalently, \( p(x) = Ce^{-\frac{1}{100}x} \). The initial condition \( p(100) = 20.09 \) gives \( 20.09 = Ce^{-1} \). Solving for \( C \), we find \( C = 20.09e \). Thus, \( p(x) = 20.09e^{(1 - \frac{x}{100})} \).

Step 2: Calculate the Unit Prices for 10 and 90 Units

Using the expression for \( p(x) \) from Step 1, we need to find \( p(10) \) and \( p(90) \).For a 10-unit order:\[ p(10) = 20.09e^{(1 - \frac{10}{100})} = 20.09e^{0.9} \]For a 90-unit order:\[ p(90) = 20.09e^{(1 - \frac{90}{100})} = 20.09e^{0.1} \]

Step 3: Check if Revenue is Maximized at x=100

The revenue function is given by \( r(x) = x \cdot p(x) = x \cdot 20.09e^{(1 - \frac{x}{100})} \). To find the maximum revenue, we take the derivative of \( r(x) \) with respect to \( x \), and set it to zero.\( \frac{d}{dx}[x \cdot 20.09e^{(1 - \frac{x}{100})}] = 20.09e^{(1 - \frac{x}{100})} - \frac{x \cdot 20.09e^{(1 - \frac{x}{100})}}{100} = 0 \).Simplifying, we get:\( 1 - \frac{x}{100} = 0 \) leading to \( x = 100 \).Thus, the revenue is maximized at \( x = 100 \).

Step 4: Graph the Revenue Function

To graph the revenue function \( r(x) = x \cdot 20.09e^{(1 - \frac{x}{100})} \), we plot the function over the interval \( 0 \leq x \leq 200 \) using a graphing tool. The plot should show the revenue rising, reaching a peak at \( x = 100 \), then gradually declining. This graphical representation confirms the calculations in Step 3.

Key Concepts

Continuous Discount Functions

Continuous discount functions are essential in pricing strategies as they allow a gradual reduction in the price of a product based on the quantity purchased. In the given exercise, the discount function is expressed through a differential equation, providing a dynamic framework that models how unit price changes with the number of units ordered.

Here's what a continuous discount function does:

• A discount decreases consistently with each additional unit.
• The rate of change can be constant, as shown by the equation \( \frac{d p}{d x} = -\frac{1}{100} p \).
• This mathematical approach helps in understanding price variations over a range of quantities.

In practice, continuous discount functions can optimize sales strategies by incentivizing bulk purchases while maintaining structured pricing that aligns with a company's revenue goals.

Revenue Maximization

Revenue maximization is a crucial goal for businesses as it influences profitability and sustainable growth strategies. The exercise explores this concept by questioning if the applied discounts maximize revenue for a specific quantity of orders, specifically 100 units in this context.

To determine if the firm is maximizing revenue at 100 units:

• The revenue function \( r(x) = x \cdot p(x) \) is constructed, which combines the ordered quantity and the unit price.
• Optimizing this function involves taking its derivative with respect to \( x \) and setting it to zero to find the critical points.
• Through this calculus-based approach, it's shown that \( x = 100 \) indeed maximizes revenue, ensuring the firm's pricing strategy aligns with revenue goals.

Understanding revenue maximization ensures that pricing is not overly aggressive, which could lead to reduced revenue despite higher sales volume.

Separable Differential Equations

Separable differential equations are a powerful tool for solving various mathematical problems, including those found in economics, physics, and engineering. This type of equation allows variables to be separated and integrated individually. The exercise at hand uses a separable differential equation to model the pricing function \( p(x) \).

Solving separable differential equations involves the following steps:

• Separate the variables so that each side of the equation contains only one variable (e.g., \( \frac{dp}{p} = -\frac{1}{100} dx \)).
• Integrate both sides to find a general solution, using the anti-derivative.
• Apply initial conditions to solve for constants and obtain a specific solution, such as \( p(x) = Ce^{(1 - \frac{x}{100})} \).

This method is particularly useful in deriving functions that depict real-world phenomena, using mathematical models to make informed decisions. By understanding separable differential equations, one can characterize dynamic systems efficiently and accurately.