+1124 An Airliner runs a special:
Usually the tickets are R4000 each per person. For every person more than 50, each ticket (for all passengers), will reduce by R50.
1) Calculate the ticket costs for each passenger if there are 60 passengers.
2) If x reflects the passengers over 50, write an expression for the price of a ticket
3) Show that the airliners income for a flight is: R=200 000 + 1500x - 50x^2
4) Calculate how many passengers will provide for a maximum income.
thank you all very much!
+128475 1) Ticket cost for 60 passengers = 4000 - (60 - 50)(50) = 4000 - 10(50) = R3500
2) Price of ticket for over 50 = R [ 4000 - 50x ]
3) The revenue, R = number of passengers * cost of a ticket
(50 + x)(4000 - 50x) = 20000 + 4000x - 2500x - 50x^2 = 200000 + 1500x - 50x^2
4) In the form ax^2 + bx + c
The number of passengers that will provide the max revenue is -b/[2a]
So we have
R = -50x^2 + 1500x + 200000
So......the number of passengers providing the max revenue is :
(-1500) / (2 * -50) = -1500/ -100 = 15 passengers = 15 over 50 = 65 passengers
EDIT TO CORRECT A SILLY ERROR !!!
+1124 CPhill,
thank you very very much!!..please note I posted this question again under my username, to be able to track in future, I tried to remove this post after I re-posted, but when I came back to try and remove this post, (I don't even know if it can be done)..I saw someone was busy compiling an answer. so I left it. please forgive me if I had broken any rules...
thank you so much for the answer!!
+1124 CPhill,
the final answer is 1500/100...and you say it's 150?...I know the answer can impossibly be 15, so how?...please explain, sorry for asking again..thank you kindly..
+118609 Chris made just a little error think.
4) Revenue = -50x^2 + 1500x + 200000
R = -50x^2 + 1500x + 200000
This is a concave down parabola. The maximum will be at the turning point. (the vertex)
Normally I'd use calculus for this but I'm not sure if you have done any calculus yet.
The axis of symmetry is x= -b/2a (from the quadratic equation)
x= -1500/(2*-50) = 1500/100 = 15
So for the most profit there should be 50+15 = 65 passengers.
