Math
posted by Amanda .
a bus company has 4000 passengers daily, each paying a fare of $2. for each $0.15 increase, the company estimates that it will lose 40 passengers per day. if the company needs to take in 10 450$ per day to stay in business, what fare should be charged?

increase fare by n 15 cent units
new fare = (2.00 + .15 n)
new passengers = (4000  40 n)
so
10450 = (2.00+.15n)(400040n)
10450 = 8000 + 520 n  6 n^2
so
6 n^2  520 n + 2450 = 0
3 n^2  260 n + 1225 = 0
n = [260 +/ sqrt(6760014700)]/6
n = [260 +/ sqrt(52900)]/6
n = [260 +/ 230]/6
n = [30]/6 or [490]/6
n = 5 or 81 2/3
well
2.00 + .15(5) = $2.75
which is a reasonable fare
however
2.00 +.15(81 2/3) = 14.25
and 4000  40 (81 2/3) = 733 passengers
Well, I guess not many people would ride for $14.25 per day so we better stick with the 75 cent fare increase
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