Problem 39 - College Algebra and 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

After you solve the system of equations, you will obtain the values for \(p_{1}\) and \(p_{2}\) for which the retail outlet will maximize its total revenue since we're using the demand functions and applying maximisation conditions.

Step 1: Write down the revenue function

The given functions are the demand functions for each of the products, \(x_{1}=1000-2 p_{1}+p_{2}\) and \(x_{2}=1500+2 p_{1}-1.5 p_{2}\). The total revenue \(R\) is given by \(R=x_{1} p_{1}+x_{2} p_{2}\). Substituting the demand functions into the revenue expression we get \(R=(1000-2 p_{1}+p_{2})p_{1} + (1500+2 p_{1}-1.5 p_{2})p_{2}\)

Step 2: Partially differentiate the revenue function with respect to \(p_{1}\) and \(p_{2}\)

In order to find maximum, we take the partial derivatives of the function with respect to \(p_{1}\) and \(p_{2}\) and set both derivatives equal to zero. As a result we get two equations which form a system of equations.

Step 3: Solve the resulting system of equations

After we set the derivatives equal to zero we have to solve system of equations. We can use substitution method, elimination method or any other method to solve the algebraic system of equations. After we solve we'll get the values for \(p_{1}\) and \(p_{2}\).