With a fare of $1.5 there are 4000 passengers of the bus service.
A $0.25 increase in fare decreases the number of passengers by 100 and a $0.25 decrease in fare increases the number of passengers by 100.
For a change in the fare of x, the total revenue
R = (4000 - 100*x)(1.5 + x*0.25)
=> 6000 - 150x + 1000x - 25x^2
To maximize R, solve R' = 0
R' = -150 + 1000 - 50x
-150 + 1000 - 50x = 0
=> 50x = 850
=> x = 17
The price of ticket should be increased by $4.25 to $5.75
Revenue is maximized when the fare is increased to $5.75.