### Maximum Revenue Optimization (Optimizing Functions with Several Variables) **Problem Statement:** An airline flying to a Midwest destination can sell 20 coach-class tickets per day at a price of $250 and six business-class tickets per day at a price of $750. It finds that for each $10 decrease in the price of the coach ticket, it will sell four more per day, and for each $50 decrease in the business-class price, it will sell two more per day. **Questions:** a) What prices should the airline charge for the coach- and business-class tickets to maximize revenue? b) How many of each type will be sold at these prices? c) What is the daily maximum revenue? **Hint:** Let \( x \) be the number of $10 price decreases for coach tickets and \( y \) be the number of $50 price decreases for business-class tickets. **Approach:** 1. **Define the Variables:** - Let \( x \) be the number of $10 price decreases for coach tickets. - Let \( y \) be the number of $50 price decreases for business-class tickets. 2. **Formulate the Revenue Equations:** - New price of coach ticket: \( 250 - 10x \) - New price of business-class ticket: \( 750 - 50y \) - Number of coach tickets sold per day: \( 20 + 4x \) - Number of business-class tickets sold per day: \( 6 + 2y \) - Revenue from coach tickets: \( (250 - 10x) \times (20 + 4x) \) - Revenue from business-class tickets: \( (750 - 50y) \times (6 + 2y) \) 3. **Total Revenue Function:** \[ R(x, y) = (250 - 10x)(20 + 4x) + (750 - 50y)(6 + 2y) \] 4. **Maximize the Total Revenue Function:** - Use calculus to take the partial derivatives of the revenue function with respect to \( x \) and \( y \). - Set the partial derivatives equal to zero and solve for \( x \) and \( y \). **Solution Steps:** 1. **Partial Derivatives and Critical Points:** \[ \frac{\