Math CentralQuandaries & Queries


Question from karen, a parent:

a charter plane company advertises that it will provide a plane for a fare of $60. if your party is twenty or less and all passengers will receive a discount of $2 per person if the party is greater than 20. what number of passengers will maximize revenue for the company

Hi Karen. The revenue is 60p for p ≤ 20, so that means revenue climbs as the number of passengers climbs: 60, 120, 180, ..., 1200 for 20 passengers.

If they carry a load of more than 20 then it continues with 58p while p > 20. So 21 passengers is 21*58 = 1218 and so it would be 1276 for 22 people etc. until you reach the capacity of the plane.

So what is the "Maximum Revenue" in this situation? It appears to increase always. So the maximum revenue is a full plane with all the passengers possible.

Are you sure you have worded the problem correctly?

Stephen La Rocque

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