Walter H. answered • 03/28/17
Tutor
New to Wyzant
10 Year Veteran Math Teacher
I think this question might need rewording.
If 75 passengers pay $225 each, that is $16875. The airline is going to get MORE passengers to LOWER its revenue? Are the additional passengers (past 75) also paying a fair? Is it the same at the first 75 passengers?
If each additional passenger after 75 gives a $5 for each passenger (but they don't pay for themselves), then the 76th passenger would decrease the revenue to $16500, and the 77th passenger would decrease the revenue to $16125. A 78th passenger would decrease the revenue to $15750.
If each additional passenger after 75 gives a $5 for each passenger (and they pay the same rate as the others), then the 76th passenger would decrease the revenue to $16720, the 77th passenger would decrease the revenue to $16555, the 78th ($16380), the 79th ($16195), and 80 ($16000).
I guess the answer would be 80, but it just seems very odd.
The way you find it mathematically is by writing your revenue function:
Rev=(cost of each passenger)(# of passengers)
The cost varies with the number of passengers past 75:
Cost=225-5(x-75)=600-5x
Multiplying by the number of passengers:
Revenue=(-5x+600)(x)=-5x^2+600x=16000
I then solve the quadratic:
-5x^2+600x-16000=0 (divide everything by -5)
x^2-120x+3200=0 (factor)
(x-40)(x-80)=0
So, x (number of passengers) has to be 40 or 80. Since 40 is less than 75, it doesn't make sense, so the answer must be 80.
If each additional passenger after 75 gives a $5 for each passenger (but they don't pay for themselves), then the 76th passenger would decrease the revenue to $16500, and the 77th passenger would decrease the revenue to $16125. A 78th passenger would decrease the revenue to $15750.
