Problem 39 - Calculus an Applied Approach

Question

Find \(p_{1}\) and \(p_{2}\) so as to maximize the total revenue \(R=x_{1} p_{1}+x_{2} p_{2}\) for a retail outlet that sells two competitive products with the given demand functions. $$ x_{1}=1000-2 p_{1}+p_{2}, x_{2}=1500+2 p_{1}-1.5 p_{2} $$

Step-by-Step Solution

Answer

The optimal prices \(p_{1}\) and \(p_{2}\) that maximize the revenue are \(p_{1} = 150\) and \(p_{2} = 200\).

Step 1: Rewrite the Revenue Equation

Rewrite the revenue equation \(R=x_{1} p_{1}+x_{2} p_{2}\) in terms of prices \(p_{1}\) and \(p_{2}\), using the given demand functions. Substitute\(x_{1}\) and \(x_{2}\) from their respective equations to get the revenue equation in terms of \(p_{1}\) and \(p_{2}\). The rewritten equation becomes \(R=(1000-2 p_{1}+p_{2}) p_{1}+(1500+2 p_{1}-1.5 p_{2}) p_{2}\).

Step 2: Differentiate R w.r.t. \(p_{1}\) and \(p_{2}\)

To find the maximum revenue, we should differentiate R with respect to \(p_{1}\) and \(p_{2}\) separately and equate each to zero to solve for \(p_{1}\) and \(p_{2}\). So, \(\frac{\partial R}{\partial p_{1}} = 0\) and \(\frac{\partial R}{\partial p_{2}} = 0\). Doing this we get two equations: \(700 - 4 p_{1} + 2 p_{2} = 0\) and \(550 + p_{1} - 1.5 p_{2} = 0\). Now we have a system of two equations that need to be solved.

Step 3: Solve the System of Equations

Solving the system of equations found in step 2, we get the optimal prices. By solving, we get \(p_{1} = 150\) and \(p_{2} = 200\). These are the prices that maximize the revenue of the two products being sold.

Key Concepts

Demand Functions

Demand functions describe how the quantity demanded of a product is influenced by various factors like price and consumers' preferences. In our exercise, two demand functions are provided:

• \( x_{1} = 1000 - 2p_{1} + p_{2} \)
• \( x_{2} = 1500 + 2p_{1} - 1.5p_{2} \)

These equations reflect the relationship between the price and demand for two competing products. Here, \( x_{1} \) and \( x_{2} \) denote the quantity demanded for each product, while \( p_{1} \) and \( p_{2} \) are their prices. Notice how changes in prices affect demand heights differently: the coefficient of \( p_{1} \) in \( x_{1} \) is \(-2\), showing that an increase in \( p_{1} \) decreases \( x_{1} \). Additionally, the cross-effects of each product on the other’s demand are captured with variables like \( p_{2} \) in \( x_{1} \) and \( p_{1} \) in \( x_{2} \). Real-world insights derived from demand functions help businesses set prices to attract more customers without losing market competitiveness.

Partial Derivatives

To find the values of \( p_{1} \) and \( p_{2} \) that maximize the revenue function \( R \), we use the technique of partial derivatives. A partial derivative measures how a function changes as one of its input variables changes while keeping the other variables constant. For the revenue function:

• The revenue \( R \) is a function of \( p_{1} \) and \( p_{2} \) based on demand: \( R = (1000 - 2p_{1} + p_{2})p_{1} + (1500 + 2p_{1} - 1.5p_{2})p_{2} \).

Partial derivatives let us assess how changes in either \( p_{1} \) or \( p_{2} \) independently impact \( R \). Calculating \( \frac{\partial R}{\partial p_{1}} \) and \( \frac{\partial R}{\partial p_{2}} \), and setting each to zero gives the conditions for a revenue maximum. This step leads to two equations that showcase relationships between variables, which are crucial for optimal decision-making in economic contexts.

System of Equations

Solving the system of equations derived from the partial derivatives enables finding the optimal prices, \( p_{1} \) and \( p_{2} \), for maximizing revenue. Once we differentiate \( R \) with respect to both \( p_{1} \) and \( p_{2} \) and equate these derivatives to zero, the result is a set of simultaneous linear equations:

• \( 700 - 4p_{1} + 2p_{2} = 0 \)
• \( 550 + p_{1} - 1.5p_{2} = 0 \)

The solution involves finding values of \( p_{1} \) and \( p_{2} \) that satisfy both equations concurrently, which are \( p_{1} = 150 \) and \( p_{2} = 200 \). Systems of equations like these are commonly used in economics to solve for multiple variables where constraints or relationships exist, helping businesses and economists predict outcomes and set strategies effectively.