SOLUTION: A harbor ferry service has about 240 000 riders per day for a fare of $2. The port authority wants to increase the fare to help with increasing operational costs. Research has show

Question 969591: A harbor ferry service has about 240 000 riders per day for a fare of $2. The port authority wants to increase the fare to help with increasing operational costs. Research has shown that for every $0.10 increase in the fare the number of riders will drop by 10 000.
a)What increase in the fare will maximize the revenue?
b) what is the new fare?
c) what is the revenue that will be received from the new fare?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A harbor ferry service has about 240 000 riders per day for a fare of $2.
The port authority wants to increase the fare to help with increasing operational costs.
Research has shown that for every $0.10 increase in the fare the number of riders will drop by 10 000.
:
a)What increase in the fare will maximize the revenue?
let x = no. of 10 cent increases and also = no. 10000 rider decreases
Revenue = riders * fare
r(x) = (240000 - 10000x)(2 + .10x)
FOIL
r(x) = 480000 + 24000x - 20000x - 1000x^2
r(x) = -1000x^2 + 4000x + 480000
simplify, divide by 1000
r(x) = -x^2 + 4x + 480
The axis of symmetry will give us max revenue; x = -b/2a
x = -4/(2*-1)
x = 2 fare increases of $.10
:
b) what is the new fare?
2 + 2(.10) = $2.20 is the new fare
:
c) what is the revenue that will be received from the new fare?
(240000 - 2(10000)) * $2.20 = $484,000
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