Maximizing profit Suppose you own a tour bus and you book groups of 20 to 70 people for a day tour. The cost per person is $30 minus $0.25 for every ticket sold. If gas and other miscellaneous costs are $200, how many tickets should you sell to maximize your profit? Treat the number of tickets as a nonnegative real number.

Solution 48E Step 1 Consider R as the profit that will be made by selling the bus tickets and N as the total number of tickets being sold, and P as the price for one person. And since the miscellaneous costs are $200 , the profit is calculated using the formula, profit = (number of tickets being sold)(cost per person) (miscellaneous cost) R = NP 200 Consider x as the increment value in the number of the tickets. For every ticket sold, the cost of the ticket decreases in $0.25 from the base price of $30. Therefore, number of tickets being sold = increment value in the number of tickets N = 1 + x And for every ticket sold cost of ticket is decreased by $0.25 from the base price of $30. That is P = 30 0.25x Substitute into , R = NP 200 =(1 + x)(30 0.25x) 200 =30+29.75x-0.25x 200 R = 170 + 29.75x 0.25x 2 Find the critical point by equating the derivative to zero. dR=0 dx d 2 dx( 170 + 29.75x 0.25x )=0 29.75 0.25(2x) = 0 29.75 0.5x = 0 x = 29.75 0.5 x = 59.5 And since R =.5 is a negative value, the profit is only maximized at this critical point. Substitute into N, N = 1 + x = 1 + 59.5 = 60.5 60 Thus the minimum number of tickets to be sold to maximize the profit is 60.