SOLUTION: a charter company will provide for a fare of $60 each for 20 or fewer passengers. for each passenger in excess of 20, the fare is decreased $2 per person for everyone what number o
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Question 259462: a charter company will provide for a fare of $60 each for 20 or fewer passengers. for each passenger in excess of 20, the fare is decreased $2 per person for everyone what number of passengers will produce the greatest revenue for the company
OHH I REALLY NEED HELP WITH THIS!PLEASE!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
a charter company will provide for a fare of $60 each for 20 or fewer passengers.
for each passenger in excess of 20, the fare is decreased $2 per person for everyone what number of passengers will produce the greatest revenue for the company
:
Let x = no. of passengers greater than 20
:
Revenue = no. of passengers * price per passenger
r = (20+x)*(60-2x)
FOIL
r = 1200 - 40x + 60x - 2x^2
A quadratic equation
-2x^2 + 20x + 1200 = 0
:
We can find x that gives the max revenue by finding the axis of symmetry; x = -b/(2a)
In this equation a = -2; b = 20
x =
x =
x = +5
;
We can say 25 passengers will give max revenue
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