Question 1120665
It would be easier to use 1 passenger for every $3 raise.
x=number of passengers changing
(81-x)(210+3x)=17010+243x-210x-3x^2
This is -3x^2+33x+17010
x-value for maximum is at x=-b/2a or -33/6 or -5.50
This makes x 75.5, with a fare of $226.50, but we need integer number of passengers. Because this is not linear, we need to look at both sides of the decimal, at 75 and 76.
76 passengers is the optimal number at a fare of $225 each.

81---210
80---213
79---216
78---219
77---222=17094
76---225=17100
75---224=16800