Find the rate and maximum income

Question

A car rental agency rents 220 cars per day at a rate of 29 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

220 cars at $29 per car daily

215 cars at $30

210 at $31

0 at $73, no one rents at 75 dollars

365 at $0 price, when free, no one rents a car more than 365 daily

graphically the points are (29,220), (30,215), (31,210) 5 less for each $1 increase,

and intercepts (0,73), (0,365). Connect the two intercepts. Revenue maximizing point is

the midpoint of that line segment (73/2, 365/2) = (x, P)

linear demand curve

x =quantity demanded = 220 - 5(P-29) = 220+145-5P= 365-5P

x(P) = 365-5P

73/2= $36.50 daily maximizes income= (365/2)(73/2)= 26645/4=$6,661.25

income= px= 365P-5P^2

take the derivative, set =0

(px)'= 365-10P= 0

P = 365/10 = $36.50 daily charge maximizes income

max income = (36.5)(365/2) =6661.25