SOLUTION: The profit function for the Recklus Hang Gliding service is P(x)=0.4x^2+fx-m, where f represents the set up fee for a customer's daily excursion and m represents the monthly hanger

Question 338218: The profit function for the Recklus Hang Gliding service is P(x)=0.4x^2+fx-m, where f represents the set up fee for a customer's daily excursion and m represents the monthly hanger rental. Also, P represents the monthly profit in dollars of the small business where x is the number os flight excursions facilitated in that month.
If $40 is charged for a set up fee, and the monthly hanger rental is $800; write an equation for the profit, P, in terms of x
This is what I came up with
P=0.4x^2+40x-800
P=0.4(30)^2+40(30)-800
From there I don't know what to do. I don't even know if that is correct.
The rest of the problem is How much is the profit when 30 flight excursions are sold in one month?
How many flight excursions must be sold in order to maximize the profit?

Answer by katealdridge(100) (Show Source): You can put this solution on YOUR website!
This problem seems strange to me in that profit equations tend to be negative parabolas. Therefore, the vertex of the parabola is when profit is maximized. However, your equation is a positive parabola, which would lead me to believe there is some sort of error in the question. The equation you wrote seems correct based on the info given, as well as your substituting 30 for x to find the profit, given 30 excursions. However maximizing profit seems impossible, given that the parabola increases toward infinity.
If you have any further questions please check out my facebook page: Kate Calendrillo, tutor. I'd be happy to answer your questions.

Ok, so the equation is So to maximize the profit you need to find the vertex of the parabola. You can do that on a graphing calculator or by using the equation where a=-0.4 and b=40. So
This equals 50. So at 50 excursions per month, the company maximizes profits. If you want to know how much profit that is, substitute 50 for x in your profit equation. P=$200
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