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)=\square \quad (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, \( p'(t) = \frac{2000(4t + 75) - 2000t(4)}{(4t + 75)^2} \)

Simplify numerator: \( 8000t + 150000 - 8000t = 150000 \)

Thus, \( p'(t) = \frac{150000}{(4t + 75)^2} \)

Step2: Calculate \( p'(9) \)

Substitute \( t = 9 \) into \( p'(t) \):

\( p'(9) = \frac{150000}{(4(9) + 75)^2} = \frac{150000}{(36 + 75)^2} = \frac{150000}{(111)^2} \)

Calculate \( 111^2 = 12321 \)

Then \( p'(9) = \frac{150000}{12321} \approx 12.174 \)

Answer:

\( 12.174 \)