posted by ted .
The city bus company usually transports 12000 riders per day at a ticket price of $1. The company wants to raise the tickets price and knows that for every 10 cents increase, the number of riders decreases by 400.
A. What price for a ticket will maximize revenue?
n riders, q price increase over 100 pennies in pennies
n = 12000 (1 - 400 *10q)
r = n ( 100 + q)
r = 12000 (1-4000q)(100+q)
r = 12000 (100 -399,999 q - 4,000q^2)
dr/dq = 12000 (-399,999 - 8,000 q)
max when 0
q = about -50 cents
looks like they should lower the price to 50 cents to maximize revenue
If you do not do calculus find the vertex of the parabola
40 q^2 + 4000q -1 = 0