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.**