Question

A small business assumes that the demand function for one of its new products can be modeled by \(p=C e^{k x} .\) When \(p=\$ 45, x=1000\) units, and when \(p=\$ 40, x=1200\) units. (a) Solve for \(C\) and \(k\). (b) Find the values of \(x\) and \(p\) that will maximize the revenue for this product.

Short answer

The constants for the demand function are \(C = \frac{45}{e^{1000 \times ln(1.125)/200}}\) and \(k = \frac{ln(1.125)}{200}\). The values of x and p that maximize revenue can be found by finding the derivative of the revenue function, setting it to 0, and solving for x and p.

Step-by-step solution

Determine the demand function

First, substitute the given points into the equation \(p=C e^{k x}\) to get two separate equations. When \(x=1000\), \(p=45\), so the equation is \(45 = C e^{1000k}\), and when \(x=1200\), \(p=40\), so the equation is now \(40 = C e^{1200k}\).

Solve for C and k

To find the values of C and k, divide the two equations: \(\frac{45}{40} = \frac{C e^{1000k}}{C e^{1200k}}\)\nThis simplifies to \(1.125 = e^{200k}\). Solve for \(k\) by taking the natural logarithm of both sides to get \(k = \frac{ln(1.125)}{200}\).\nNow, substitute \(k\) into the first equation and solve for \(C\). This gives \(C = \frac{45}{e^{1000k}}\).

Find the revenue function

The revenue R is given by the product of the price p and the quantity x. So, \(R = p \cdot x = Cxe^{kx}\). Substitute the values for C and k found earlier.

Maximize the revenue

To maximize the revenue, take the derivative of R with respect to x and set it equal to 0. This will give you a value for x. Substitute this x value into the demand function to find the corresponding price p.

Key concepts

Demand Function

Understanding the demand function in economics is essential to predicting consumer behavior. In our context, the demand function is represented as \[ p = C e^{kx} \]where:

• \( p \): price of the product
• \( C \): constant which scales the exponential function
• \( x \): quantity of the product demanded
• \( k \): a constant representing the growth rate of the function in relation to quantity

This function suggests that the price changes exponentially with the change in quantity. For the given problem, we used two given conditions, (\( p = 45, x = 1000 \) and \( p = 40, x = 1200 \)), to find the constants \( C \) and \( k \). By substituting these conditions into the function, we set up two equations, enabling us to solve for \( C \) and \( k \) through mathematical manipulations such as division and logarithms. This is crucial, as these constants help describe the unique demand curve for the product in question.

Exponential Models

Exponential models are widely used in various fields such as finance, biology, and economics. These models are characterized by a constant relative growth rate, meaning that the growth of the function is proportional at every point in time. In our demand function, the model is exponential due to the presence of the term \[ e^{kx} \]which indicates that any change in \( x \), the quantity, results in a compounding effect on the price \( p \).
Certain characteristics of exponential functions make them suitable for modeling economic behaviors:

• Rapid growth or decay
• Predictability over continuous times
• Relationship modeling between varying variables such as price, demand, and supply

Exponential models are effective in projecting future sales trends and setting optimal pricing strategies in business scenarios. In our exercise, this exponential behavior is directly studied by evaluating how changes in unit sales \( x \) influence the price \( p \) exponentially.

Derivative Application

Applying derivatives is a fundamental technique in calculus used to find rates of change and to optimize functions. In this scenario, the derivative is used to maximize the revenue function. The formula for revenue, given as \[ R = p \cdot x = Cxe^{kx} \]is a product of price and quantity. To find the revenue's maximum point, the derivative of this function with respect to \( x \) is taken.

• Differentiate the revenue function: This involves applying the product rule and the chain rule due to the exponential component.
• Set the derivative equal to zero: This condition helps locate critical points where maximum revenue might occur.
• Verify the maximum point: Often second derivatives are used to confirm whether the critical point is a maximum.

This derivative application is crucial in determining the optimal quantity of the product that should be sold to achieve maximum revenue, guiding businesses in making informed economic decisions.