the revenue, in dollars, made by an auto repair ceenter t days after the start of the month is modeled by r(t)=900e^0.31t. the number of customers who visited the auto repair center t dyas after the start of the month is modeled by c(t)= sqaureroot240t. the revenue earned per customer is given by g(t)=r(t)/c(t). how fast is the rvenue earned per customer changing on the 9th day
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To find how fast the revenue earned per customer is changing on the 9th day, we need to calculate the rate of change of \( g(t) = \frac{r(t)}{c(t)} \) with respect to time, \( t \), and then evaluate it at \( t = 9 \).

### Step 1: Define the functions
We have the following functions:

- Revenue function: \( r(t) = 900e^{0.31t} \)
- Customer function: \( c(t) = \sqrt{240t} = (240t)^{1/2} \)
- Revenue per customer: \( g(t) = \frac{r(t)}{c(t)} \)

### Step 2: Differentiate \( g(t) \)
We use the quotient rule to differentiate \( g(t) \):
\[
g'(t) = \frac{ \frac{d}{dt}[r(t)] \cdot c(t) - r(t) \cdot \frac{d}{dt}[c(t)] }{[c(t)]^2}
\]

First, find the derivatives of \( r(t) \) and \( c(t) \).

#### Derivative of \( r(t) \):
\[
r'(t) = \frac{d}{dt}[900e^{0.31t}] = 900 \cdot 0.31e^{0.31t} = 279e^{0.31t}
\]

#### Derivative of \( c(t) \):
\[
c'(t) = \frac{d}{dt}[(240t)^{1/2}] = \frac{1}{2}(240t)^{-1/2} \cdot 240 = \frac{120}{\sqrt{240t}}
\]

### Step 3: Plug these into the quotient rule
Now substitute into the quotient rule:
\[
g'(t) = \frac{279e^{0.31t} \cdot \sqrt{240t} - 900e^{0.31t} \cdot \frac{120}{\sqrt{240t}}}{(240t)}
\]

### Step 4: Evaluate at \( t = 9 \)
We need to calculate \( g'(9) \). First, plug \( t = 9 \) into the expressions for \( r(t) \), \( r'(t) \), \( c(t) \), and \( c'(t) \).

- \( r(9) = 900e^{0.31 \times 9} = 900e^{2.79} \)
- \( r'(9) = 279e^{2.79} \)
- \( c(9) = \sqrt{240 \times 9} = \sqrt{2160} \)
- \( c'(9) = \frac{120}{\sqrt{2160}} \)

Now, substitute these values into the expression for \( g'(9) \) and simplify.

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This will give the rate at which the revenue per customer is changing on the 9th day.

Would you like me to continue simplifying this?

The revenue, in dollars, made by an auto repair center t days after the start of the month is modeled by r(t) = 900e^0.31t. The number of customers who visited the auto repair center t days after the start of the month is modeled by c(t) = √240t. The revenue earned per customer is given by g(t) = r(t) / c(t). How fast is the revenue earned per customer changing on the 9th day?

Explore how to calculate the rate of change in revenue per customer for an auto repair center on the 9th day. This problem applies calculus concepts such as the quotient rule, exponential functions, and radical functions, making it suitable for students in Grades 11-12 or early college calculus.

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