https://file.aiquickdraw.com/mathbot/file/172611732047556f2exhe.jpg

We are given the following functions in the problem:

1. The revenue, \( r(t) \), is modeled by:
   \[
   r(t) = 900e^{0.3t}
   \]
   
2. The number of customers, \( c(t) \), is modeled by:
   \[
   c(t) = \sqrt{200t}
   \]

3. The revenue per customer, \( g(t) \), is given by:
   \[
   g(t) = \frac{r(t)}{c(t)}
   \]

We are tasked with finding how fast the revenue earned per customer is changing on the 10th day, i.e., calculating \( g'(10) \).

### Step-by-step approach:

1. **Differentiate \( g(t) \):**
   To find \( g'(t) \), we need to use the quotient rule for derivatives:
   \[
   \left( \frac{f(t)}{h(t)} \right)' = \frac{f'(t)h(t) - f(t)h'(t)}{h(t)^2}
   \]
   In this case:
   \[
   f(t) = r(t) \quad \text{and} \quad h(t) = c(t)
   \]
   
2. **Differentiate \( r(t) \):**
   \[
   r'(t) = \frac{d}{dt} \left( 900e^{0.3t} \right) = 900 \cdot 0.3e^{0.3t} = 270e^{0.3t}
   \]
   
3. **Differentiate \( c(t) \):**
   \[
   c(t) = \sqrt{200t} = (200t)^{1/2}
   \]
   Using the chain rule:
   \[
   c'(t) = \frac{1}{2}(200t)^{-1/2} \cdot 200 = \frac{100}{\sqrt{200t}}
   \]

4. **Apply the quotient rule:**
   Now apply the quotient rule to find \( g'(t) \):
   \[
   g'(t) = \frac{r'(t) \cdot c(t) - r(t) \cdot c'(t)}{c(t)^2}
   \]
   
   Substituting the expressions for \( r'(t) \), \( r(t) \), \( c(t) \), and \( c'(t) \):
   \[
   g'(t) = \frac{(270e^{0.3t}) \cdot (\sqrt{200t}) - (900e^{0.3t}) \cdot \left( \frac{100}{\sqrt{200t}} \right)}{(\sqrt{200t})^2}
   \]
   
   Simplify the expression step by step.

5. **Evaluate at \( t = 10 \):**
   Substitute \( t = 10 \) into the simplified expression to find \( g'(10) \).

Let me now compute this.The rate at which the revenue earned per customer is changing on the 10th day is approximately **101.05 dollars per customer per day**.

Would you like more details on any part of this solution?

Here are 5 related questions to expand on this topic:
1. How would the result change if the growth rate of revenue, \( 0.3 \), was increased?
2. What happens to the revenue per customer if the number of customers grows at a faster rate?
3. How does the quotient rule help in calculating the derivative of revenue per customer?
4. What is the significance of the exponential function in the revenue model?
5. How can you interpret the second derivative \( g''(t) \) in this context?

**Tip:** The quotient rule is essential for calculating the derivative of functions that are ratios, like revenue per customer.

How fast is the revenue earned per customer changing on the 10th day?

Discover how to calculate the rate of change of revenue per customer on the 10th day using differentiation and the quotient rule. This calculus problem involves exponential functions and square roots, suitable for advanced high school or college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation