I interpret this question: for every $0.10 increase in the bus fare 50 fewer passengers per hour ride the bus.
We can build an equation which gives the revenue per hour R(n) for each increase of the fare. After n increases in the fare:
R(n)=(500-50n)(0.50+0.10n)=250+25n-5n²=
5(50+5n-n²).
We need to find the maximum value of this expression, so we complete the square:
5(50-(n²-5n+6.25-6.25))=
5(50-(n-2.5)²+6.25)=
250+31.25-(n-2.5)²=
281.25-(n-2.5)².
When n=2.5, corresponding to an increase in the fare of $0.25 to 0.50+0.25=$0.75, and a reduction in passengers per hour of 125 to 500-125=375, the revenue is maximised to $281.25. This means the revenue is 375×0.75=$281.25 and the fare cost is $0.75.