Question

optimization: problem 2
(1 point)
a car rental agency rents 220 cars per day at a rate of 31 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
note: you can earn partial credit on this problem.

Explanation:

Step1: Define variables and income function

Let \( x \) be the number of 1 - dollar increases.
The rate per car is \( r=(31 + x) \) dollars.
The number of cars rented is \( n=(220-5x) \).
The income \( I(x)=(31 + x)(220 - 5x) \).
Expand \( I(x)=31\times220-31\times5x+220x-5x^{2}=6820 + 65x-5x^{2} \).

Step2: Find the vertex of the quadratic function

For a quadratic function \( y = ax^{2}+bx + c\) (\(a=-5\), \(b = 65\), \(c = 6820\)), the \(x\) - coordinate of the vertex is given by \(x=-\frac{b}{2a}\).
Substitute \(a=-5\) and \(b = 65\) into \(x =-\frac{b}{2a}\), we get \(x=-\frac{65}{2\times(-5)}=\frac{65}{10}=6.5\).

Step3: Calculate the rate and income

The rate \(r=31 + x\), substituting \(x = 6.5\), we have \(r=31+6.5 = 37.5\) dollars/day.
The income \(I(x)=-5x^{2}+65x + 6820\), substitute \(x = 6.5\) into it.
\(I(6.5)=-5\times(6.5)^{2}+65\times6.5 + 6820\)
\(=-5\times42.25+422.5+6820\)
\(=-211.25+422.5+6820\)
\(=7031.25\) dollars/day.

Answer:

Rate = \(37.5\) dollars/day
Maximum Income = \(7031.25\) dollars/day