Question

attempt 1: 10 attempts remaining. the revenue, in dollars, made by an auto - repair center ( t ) days after the start of the month is modeled by ( r(t)=650e^{0.33t} ). the number of customers who visited the auto - repair center ( t ) days after the start of the month is modeled by ( c(t)=sqrt{250t} ). the revenue earned per customer is given by ( g(t)=\frac{r(t)}{c(t)} ). how fast is the revenue earned per customer changing on the 8th day? (round to the nearest cent.) dollars per customer per day submit answer

Explanation:

Step1: Find the derivative of \( g(t) \) using the quotient rule

The quotient rule states that if \( g(t)=\frac{u(t)}{v(t)} \), then \( g^\prime(t)=\frac{u^\prime(t)v(t)-u(t)v^\prime(t)}{[v(t)]^2} \). Here, \( u(t) = r(t)=650e^{0.33t} \) and \( v(t)=c(t)=\sqrt{250t}=(250t)^{\frac{1}{2}} \).

First, find \( u^\prime(t) \): The derivative of \( e^{kt} \) is \( ke^{kt} \), so \( u^\prime(t)=650\times0.33e^{0.33t}=214.5e^{0.33t} \).

Next, find \( v^\prime(t) \): Using the power rule and chain rule, \( v^\prime(t)=\frac{1}{2}(250t)^{-\frac{1}{2}}\times250=\frac{125}{\sqrt{250t}} \).

Now, apply the quotient rule:
\[
g^\prime(t)=\frac{214.5e^{0.33t}\times\sqrt{250t}-650e^{0.33t}\times\frac{125}{\sqrt{250t}}}{250t}
\]

Step2: Simplify the numerator

First, rewrite the terms with a common denominator in the numerator. The common denominator is \( \sqrt{250t} \).

\[

$$\begin{align*} \text{Numerator}&=214.5e^{0.33t}\times\sqrt{250t}\times\frac{\sqrt{250t}}{\sqrt{250t}}-650e^{0.33t}\times\frac{125}{\sqrt{250t}}\times\frac{\sqrt{250t}}{\sqrt{250t}}\\ &=\frac{214.5e^{0.33t}\times250t - 650e^{0.33t}\times125}{ \sqrt{250t}}\\ &=\frac{e^{0.33t}(214.5\times250t - 650\times125)}{\sqrt{250t}}\\ &=\frac{e^{0.33t}(53625t - 81250)}{\sqrt{250t}} \end{align*}$$

\]

So, \( g^\prime(t)=\frac{\frac{e^{0.33t}(53625t - 81250)}{\sqrt{250t}}}{250t}=\frac{e^{0.33t}(53625t - 81250)}{250t\sqrt{250t}} \)

Step3: Evaluate \( g^\prime(t) \) at \( t = 8 \)

First, calculate the values of \( t = 8 \) in the expressions:

• Calculate \( e^{0.33\times8}=e^{2.64}\approx14.09 \)
• Calculate \( 53625\times8 - 81250=429000 - 81250 = 347750 \)
• Calculate \( 250\times8=2000 \) and \( \sqrt{250\times8}=\sqrt{2000}\approx44.72 \)

Now substitute into \( g^\prime(t) \):

\[

$$\begin{align*} g^\prime(8)&=\frac{14.09\times347750}{2000\times44.72}\\ &=\frac{14.09\times347750}{89440}\\ &\approx\frac{4900000}{89440}\\ &\approx54.8 \end{align*}$$

\]

Answer:

The revenue per customer is changing at approximately \(\boxed{54.8}\) dollars per customer per day on the 8th day.