# Math

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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?

• Math -

increase fare by n 15 cent units
new fare = (2.00 + .15 n)
new passengers = (4000 - 40 n)
so
10450 = (2.00+.15n)(4000-40n)
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(67600-14700)]/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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