A rail company, QNRail, is selling tickets for a special trip from Toronto to Winnipeg. The train’s maximum capacity is 200 passengers. All tickets are sold at the same price p in Canadian dollars. Based on market research, they know that the quantity q of tickets demanded depends on the price p charged per ticket, according to the formula below.
q = (300 − 2p if p < 100
250 − 3/2 p if p ≥ 100)
In order to maximize their revenue for this trip, how much should QNRail charge per ticket? Please show all of your reasoning! Round your final answer to the nearest cent.
I speak about the first equation, the principle is the same for the second one.
$300 − 2p$ represents the number of tickets sold at a certain price.
So $(300 − 2p) * p$ represents the revenue.
From the derivative $300 - 4p$ you can see the maximum at $p = 75$
Putting this value in the function $(300 − 2p) * p$ you find the maximum revenue for a price $p < 100$
Do the same for the other case and compare the results.