Question

attempt 1: 10 attempts remaining. the daily revenue at the splashzone water park is modeled by $r(p)=-0.02p^{3}+2.40p^{2}-0.75p$, where $r$ is in dollars and $p$ is the number of patrons on that day. the park opens may 1st and stays open for 154 days. the number of patrons grows according to $p(t)=0.83t + 38.00$, where $t$ is the number of days since the park opened. a) find the revenue as a function of the number of days since the park opened, $r(p(t))$: (enter your answer with all coefficients to 2 decimal places, e.g., $-0.02(0.75t + 40.00)^{3}$) b) find $r(t)$ when $t = 84$ days, $r(84)$: (round to two decimal places as needed.)

Explanation:

Step1: Substitute $p(t)$ into $R(p)$

Substitute $p = 0.83t + 38.00$ into $R(p)=- 0.02p^{3}+2.40p^{2}-0.75p$.
\[

$$\begin{align*} R(p(t))&=-0.02(0.83t + 38.00)^{3}+2.40(0.83t + 38.00)^{2}-0.75(0.83t + 38.00) \end{align*}$$

\]
First, expand $(a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}$ and $(a + b)^{2}=a^{2}+2ab + b^{2}$ where $a = 0.83t$ and $b = 38.00$.
\[

$$\begin{align*} (0.83t+38.00)^{3}&=(0.83t)^{3}+3\times(0.83t)^{2}\times38.00 + 3\times0.83t\times38.00^{2}+38.00^{3}\\ &=0.571787t^{3}+3\times0.6889t^{2}\times38.00+3\times0.83t\times1444 + 54872\\ &=0.571787t^{3}+78.5346t^{2}+3591.24t + 54872 \end{align*}$$

\]
\[

$$\begin{align*} (0.83t + 38.00)^{2}&=(0.83t)^{2}+2\times0.83t\times38.00+38.00^{2}\\ &=0.6889t^{2}+63.08t + 1444 \end{align*}$$

\]
\[

$$\begin{align*} R(p(t))&=-0.02(0.571787t^{3}+78.5346t^{2}+3591.24t + 54872)+2.40(0.6889t^{2}+63.08t + 1444)-0.75(0.83t + 38.00)\\ &=- 0.01143574t^{3}-1.570692t^{2}-71.8248t - 1097.44+1.65336t^{2}+151.392t+3465.6-0.6225t - 28.5\\ &=-0.01t^{3}+0.08t^{2}+78.95t + 2339.66 \end{align*}$$

\]

Step2: Differentiate $R(p(t))$

Differentiate $R(p(t))=-0.01t^{3}+0.08t^{2}+78.95t + 2339.66$ with respect to $t$.
Using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have:
$R'(t)=-0.01\times3t^{2}+0.08\times2t + 78.95$
$R'(t)=-0.03t^{2}+0.16t + 78.95$

Step3: Evaluate $R'(t)$ at $t = 84$

Substitute $t = 84$ into $R'(t)$.
$R'(84)=-0.03\times84^{2}+0.16\times84 + 78.95$
$R'(84)=-0.03\times7056+13.44 + 78.95$
$R'(84)=-211.68+13.44 + 78.95$
$R'(84)=-119.29$

Answer:

a) $R(p(t))=-0.01t^{3}+0.08t^{2}+78.95t + 2339.66$
b) $-119.29$