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 mothly hanger
Algebra ->
Graphs
-> 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 mothly hanger
Log On
Question 229253: 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 mothly hanger rental. Also, Prepresents the monthly profit in dollars of the small business where x is the number of flight excursions facilitated in that month. If $40 is charged for a set up fee, and the monthly hanger is $800. How many flight excursions must be sold in order to maximize the profit and what is the maximum profit? Trial and error is not appropriate method. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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 of flight excursions facilitated in that month.
If $40 is charged for a set up fee, and the monthly hanger is $800.
How many flight excursions must be sold in order to maximize the profit and what is the maximum profit? T
:
Using the given value in the equaton
P(x) = -.4x^2 + 40x - 800
:
Maximum occurs when x = the axis of symmetry
:
We can find this using x = -b/(2a)
In this problem b=40; a=-.4
x =
x =
x = +50 flight excursions in the month
;
Find the actual profit, replace x with 50
P = -.4(50^2) + 40(50) - 800
P = -.4(2500) + 2000 - 800
P = -1000 + 2000 - 800
P = $200; not much really
