SOLUTION: Could you please help with this problem? The profit function for the recklus Hang gliding service is P(x) = -0.2x^2 +fx - m, where f represents the set up fee for a customer's

Question 462296: Could you please help with this problem?
The profit function for the
recklus Hang gliding service is P(x) = -0.2x^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 of flight excursions facilitated in a month.
A) if $30 is charged for a set up fee, and the monthly hanger rental is $600, write an equation for the profit, P, in terms of x.
b) How much is the profit when 40 flight excursions are sold in a month?
c) How many flight excursions must be sold in order to maximize the profit? Show the work algebraically.
d) What is the maximum profit.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The profit function for the
recklus Hang gliding service is P(x) = -0.2x^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 of flight excursions facilitated in a month.
A) if $30 is charged for a set up fee, and the monthly hanger rental is $600, write an equation for the profit, P, in terms of x.
b) How much is the profit when 40 flight excursions are sold in a month?
c) How many flight excursions must be sold in order to maximize the profit? Show the work algebraically.
d) What is the maximum profit.
======================================================
A) P(x) = -0.2x^2 + 30x - 600
B) P(40) = -0.2*40^2 + 30*40 - 600 = -320 + 1200 - 600 = $280
C) To maximize profit, set dP(x)/dx = 0, solve for x:
dP/dx = -0.4x + 30 = 0 -> x = -30/-0.4 = 75
D) Maximum profit is P(75) = -0.2(75)^2 + 30(75) - 600 = $525
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