Question

optimization: problem 2
(1 point)
a car rental agency rents 200 cars per day at a rate of 30 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.
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Explanation:

Step1: Define variables

Let \(x\) be the number of \(\$1\) increases in the daily rate.
The daily rate \(r=(30 + x)\) dollars/day.
The number of cars rented \(n=(200-5x)\).
The income \(I = r\times n=(30 + x)(200 - 5x)\).

Step2: Expand the income function

Using the FOIL method:

$$
LATEXBLOCK0
$$

Step3: Find the vertex of the quadratic function

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-5\), \(b = 50\), \(c = 6000\)), the \(x\)-coordinate of the vertex is given by \(x=-\frac{b}{2a}\).

$$ x=-\frac{50}{2\times(-5)}=\frac{-50}{-10} = 5 $$

Step4: Find the optimal rate and maximum income

The optimal rate \(r=30 + x\), substituting \(x = 5\), we get \(r=30+5=\$35\)/day.
The maximum income \(I=-5x^{2}+50x + 6000\), substituting \(x = 5\):

$$
LATEXBLOCK1
$$

Answer:

Rate \( = 35\) dollars/day, Maximum income \(=6125\) dollars/day