Question

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

Explanation:

Step1: Define variables

Let \(x\) be the number of 1 - dollar increases. The daily rate \(r=(30 + x)\) dollars, and the number of cars rented \(n=(200-5x)\)

Step2: Form income function

Income \(I = r\times n=(30 + x)(200 - 5x)\)
Expand: \(I=6000-150x + 200x-5x^{2}=6000 + 50x-5x^{2}\)

Step3: Find maximum of quadratic function

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-5\), \(b = 50\), \(c = 6000\)), the vertex \(x=-\frac{b}{2a}\)
\(x=-\frac{50}{2\times(-5)} = 5\)

Step4: Calculate rate and income

Rate: \(r=30 + x=30+5 = 35\) dollars/day
Income: \(I=(30 + 5)(200-5\times5)=35\times175 = 6125\) dollars/day

Answer:

Rate = \(35\) dollars/day
Maximum Income = \(6125\) dollars/day