Problem 40 - 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. $$ x_{1}=1000-4 p_{1}+2 p_{2}, x_{2}=900+4 p_{1}-3 p_{2} $$

Step-by-Step Solution

Answer

To find the values of \(p_{1}\) and \(p_{2}\) that maximize total revenue, we would solve the partial derivatives with respect to \(p_{1}\) and \(p_{2}\) for zero. The solution to this system of equations will provide the optimal prices \(p_{1}\) and \(p_{2}\).

Step 1: Substitute the demand functions

Substitute the demand functions \(x_{1}=1000-4 p_{1}+2 p_{2}\) and \(x_{2}=900+4 p_{1}-3 p_{2}\) into the revenue equation to express \(R\) as a function of \(p_{1}\) and \(p_{2}\). This gives us: \(R=(1000-4 p_{1}+2 p_{2})p_{1}+(900+4 p_{1}-3 p_{2})p_{2}\)

Step 2: Simplify the revenue function

Expand the bracket and simplify the equation from the previous step to get a quadratic function of \(p_{1}\) and \(p_{2}\): \(R=1000p_{1} -4p_{1}^{2}+2p_{1}p_{2} +900p_{2}+4p_{1}p_{2} -3p_{2}^{2}\)

Step 3: Calculate the partial derivatives

To find the maximum revenue, we assume the partial derivatives with respect to \(p_{1}\) and \(p_{2}\) to be zero. We differentiate \(R\) with respect to \(p_{1}\) and \(p_{2}\) to obtain two equations: \(\frac{\partial R}{\partial p_{1}} =0\) and \(\frac{\partial R}{\partial p_{2}} =0\) which becomes \(1000-8p_{1}+2p_{2}+4p_{2}=0\) and \(900+4p_{1}-6p_{2}=0\) upon simplification.

Step 4: Solve the system of equations

Solve the system of equations derived from setting the partial derivatives to zero. The solution of this system will give the optimal values of \(p_{1}\) and \(p_{2}\).