Question

the revenue, in dollars, made by an auto - repair center t days after the start of the month is modeled by $r(t)=800e^{0.34t}$. the number of customers who visited the auto - repair center t days after the start of the month is modeled by $c(t)=sqrt{180t}$. 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 9th day? (round to the nearest cent.)

Explanation:

Step1: Write the function $g(t)$

We know $r(t)=800e^{0.34t}$ and $c(t)=\sqrt{180t}=(180t)^{\frac{1}{2}}$, so $g(t)=\frac{800e^{0.34t}}{\sqrt{180t}}=\frac{800e^{0.34t}}{(180t)^{\frac{1}{2}}}$.

Step2: Use the quotient - rule to find $g^\prime(t)$

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 800e^{0.34t}$, so $u^\prime=800\times0.34e^{0.34t}=272e^{0.34t}$, and $v=(180t)^{\frac{1}{2}}$, so $v^\prime=\frac{1}{2}(180t)^{-\frac{1}{2}}\times180=\frac{90}{\sqrt{180t}}$. Then $g^\prime(t)=\frac{272e^{0.34t}\sqrt{180t}-800e^{0.34t}\times\frac{90}{\sqrt{180t}}}{180t}$.

Step3: Simplify $g^\prime(t)$

\[

$$\begin{align*} g^\prime(t)&=\frac{272e^{0.34t}\sqrt{180t}-\frac{72000e^{0.34t}}{\sqrt{180t}}}{180t}\\ &=\frac{\frac{272e^{0.34t}\times180t - 72000e^{0.34t}}{\sqrt{180t}}}{180t}\\ &=\frac{272e^{0.34t}\times180t - 72000e^{0.34t}}{180t\sqrt{180t}}\\ &=\frac{e^{0.34t}(48960t - 72000)}{180t\sqrt{180t}} \end{align*}$$

\]

Step4: Evaluate $g^\prime(t)$ at $t = 9$

First, substitute $t = 9$ into the formula:
\[

$$\begin{align*} g^\prime(9)&=\frac{e^{0.34\times9}(48960\times9-72000)}{180\times9\sqrt{180\times9}}\\ &=\frac{e^{3.06}(440640 - 72000)}{1620\sqrt{1620}}\\ &=\frac{e^{3.06}\times368640}{1620\times\sqrt{1620}}\\ \end{align*}$$

\]
Since $e^{3.06}\approx21.37$, $\sqrt{1620}\approx40.25$, we have:
\[

$$\begin{align*} g^\prime(9)&=\frac{21.37\times368640}{1620\times40.25}\\ &=\frac{7878836.8}{65205}\\ &\approx120.83 \end{align*}$$

\]

Answer:

$120.83$