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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.

jj889
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1 Answers1

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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.

anotherOne
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