A car rental agency rents 180 cars per day at a rate of 28 dollars per day.

Question

A car rental agency rents 180 cars per day at a rate of 28 dollars per day. For each 1 dollar increase in the daily rate, 5 fewer cars are rented. At what rate should the cars be rented to produce the maximum income, and what is the maximum income? 

Rate= ______dollars/day

Maximum income= ______dollars/day

Answer 1

Revenue = R = Px = 180(28)

180 cars at $28 per day

x drops 5, for a $1 increase in P

x=-5P+b

180=-5(28) = b

b=320

x=-5P+320

graph it, y intercept = (0,64) when no one rents at all

x intercept = (320,0) when they can't rent more, just giving car rentals for free

take average, the midpoint of the line segment connecting those two points: M(160,32)

revenue maximizing price is $32 per day

renting 160 cars for max revenue of 32(160) = $5,120 per day

that's the geometric approach with a graph of a right triangle

calculus approach is take the derivative of revenue and set it equal to zero

R= Px = -5P^2 +320P

R' = -10P +320=0

P = $32 per day

R=Px = -5(32)^2 +320(32) = $5,120 = max revneue per day