Question

the function $p(t) = \frac{2000t}{4t + 75}$ gives the population $p$ of deer in an area after $t$ months.
a) find $p(9)$, $p(45)$, and $p(90)$
b) find $p(9)$, $p(45)$, and $p(90)$
c) interpret the meaning of your answers to part (a) and (b) what is happening to the population of deer in the long term?

a) $p(9) = 12.174$ deer/month
(type an integer or decimal rounded to three decimal places as needed )
$p(45) = 2.307$ deer/month
(type an integer or decimal rounded to three decimal places as needed )
$p(90) = \square$
(type an integer or decimal rounded to three decimal places as needed )

Explanation:

Step1: Find the derivative of \( p(t) \)

We use the quotient rule: if \( p(t)=\frac{u(t)}{v(t)} \), then \( p'(t)=\frac{u'(t)v(t)-u(t)v'(t)}{v(t)^2} \). Here, \( u(t) = 2000t \), so \( u'(t)=2000 \); \( v(t)=4t + 75 \), so \( v'(t)=4 \).
\[

$$\begin{align*} p'(t)&=\frac{2000(4t + 75)-2000t(4)}{(4t + 75)^2}\\ &=\frac{8000t+150000 - 8000t}{(4t + 75)^2}\\ &=\frac{150000}{(4t + 75)^2} \end{align*}$$

\]

Step2: Evaluate \( p'(90) \)

Substitute \( t = 90 \) into \( p'(t) \):
\[

$$\begin{align*} p'(90)&=\frac{150000}{(4\times90 + 75)^2}\\ &=\frac{150000}{(360+75)^2}\\ &=\frac{150000}{435^2}\\ &=\frac{150000}{189225}\\ &\approx0.793 \end{align*}$$

\]

Answer:

\( p'(90)\approx\boxed{0.793} \) deer/month