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 )
$p(90) = 0.793$ deer/month
(type an integer or decimal rounded to three decimal places as needed )
b) $p(9) = \square$
(type an integer or decimal rounded to three decimal places as needed )

Explanation:

Step1: Find the first derivative \( p'(t) \)

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, \( u(t) = 2000t \), \( u'(t)=2000 \); \( v(t)=4t + 75 \), \( v'(t)=4 \). So,
\[
p'(t)=\frac{2000(4t + 75)-2000t(4)}{(4t + 75)^2}=\frac{8000t+150000 - 8000t}{(4t + 75)^2}=\frac{150000}{(4t + 75)^2}
\]

Step2: Find the second derivative \( p''(t) \)

Differentiate \( p'(t)=\frac{150000}{(4t + 75)^2}=150000(4t + 75)^{-2} \) using the chain rule. Let \( u = 4t+75 \), then \( p'(t)=150000u^{-2} \), \( p''(t)=150000\times(-2)u^{-3}\times4=\frac{-1200000}{(4t + 75)^3} \)

Step3: Calculate \( p''(9) \)

Substitute \( t = 9 \) into \( p''(t) \):
\[
p''(9)=\frac{-1200000}{(4\times9 + 75)^3}=\frac{-1200000}{(36 + 75)^3}=\frac{-1200000}{111^3}=\frac{-1200000}{1367631}\approx - 0.877
\]

Answer:

\( p''(9)\approx - 0.877 \)