For spring training the Diamondbacks have 1032 seats at 2600

For spring training, the Diamondbacks have 1032 seats at $26.00 at Talking Stick Stadium. For every $0.52 increase in price, 11 seats will be empty. What should the price be to maximize revenue? What is the maximum revenue? There are three ways to solve this problem: Brute force using an Excel spreadsheet Using Excel Solver Finding the vertex of a parabola Using calculus

Solution

Let R be the revenue and let the price of tickets at the Talking Stick stadium be $ (26+ x ). We know that the number of vacant seats increases by 11 for every $ 0.52 increase in ticket price. Thus, if the ticket price increases by $ x, the number of vacant seats is (11/0.52) x .Then R = Ticket price * ( number of occupied seats) = (26 +x) [ 1032- (11/0.52)x ] = (1/0.52)[ (26 + x)(536.64- 11x) = (1/0.52)[ -11x² - 286x + 536.64 x + 13592.64) or, R = ( 1/0.52)[ - 11x² + 250.64x + 13592.64] ....(1)

Now, we know that R is maximum when dR/dx = 0 and d²R/dx² is negative. Here, dR/dx = 1/(0.52)[ -22x + 250.64] and d²R/dx² = 1/(0.52) ( -22) so that if dR/dx = 0, then 1/(0.52)[ -22x + 250.64] = 0 or, -22x + 250.64 = 0 so that x = 250.64/22 = 11.39272727... = 11.39( on rounding off to 2 decimal places). Also, at this value of x, d²R/dx² is negative ( d²R/dx² is negative regardless of the value of x). Thus, R is maximum when x = 11.39. The price of the ticket to maximise revenue is $ 26 + x = $ 26 + 11.39 = $ 37.29. Also, the maximum revenue is (1/0.52)[ - 11(11.39)²+ (250.64)11.39 + 13592.64] = (1/0.52)[ -1427.05 + 2854.79 + 13592.64] = (1/0.52)( 15020.38) = $ 28885.35 ( on rounding off to the nearest cent) . We can verify the result by taking x = 11.37 and x = 11.41( to neutralize the impact of rounding off) . If x = 11.37, then R = (1/0.52)[ - 1422.05 + 2849.78 + 13592.64] = (1/0.52)(15020.37)= $ 28885.33. Also, when x = $11.41, then R = (1/0.52)[ - 1432.07 + 2859.80 + 13592.64] = (1/0.52)( 15020.37) = $28885.33. Thus, for maximizing revenue, the ticket price should be $ 37.29. Also, the maximum revenue is $ 28885.35.