Question

the function p(t) = \frac{2000t}{41 + 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?
c) what is the meaning of the answers from parts (a) and (b)?
a. the population is shrinking at a steady rate
b. the population is shrinking, but the rate of decline is slowing down
c. the population is growing, but the rate of growth is slowing down
d. the population is shrinking, and the rate of decline is increasing
e. the population is growing, and the rate of growth is increasing
f. the population is growing at a steady rate
g. the population is holding steady
what is happening to the population of deer in the long term? select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. the population of deer approaches a maximum of \square
b. the population of deer continues to grow without bound
c. the population of deer reaches a maximum of \square and then starts to decrease

Explanation:

Part (c) - Meaning of Answers

• For the first part of (c), we analyze the derivative \( p'(t) \) (rate of population change) and second derivative \( p''(t) \) (rate of change of the rate). The function \( p(t)=\frac{2000t}{41 + 75} \) (simplify denominator: \( 41+75 = 116 \), so \( p(t)=\frac{2000}{116}t=\frac{500}{29}t \)? Wait, no, maybe a typo, likely \( p(t)=\frac{2000t}{41 + 75^t} \) (logistic growth-like). Assuming logistic, \( p'(t)>0 \) (growing) and \( p''(t) \) negative (growth rate slowing). So option C: "The population is growing, but the rate of growth is slowing down" matches.
• For long - term: As \( t

ightarrow\infty \), if \( p(t)=\frac{2000t}{41 + 75^t} \), the denominator \( 75^t \) dominates, so \( p(t)
ightarrow0 \)? Wait, no, maybe original function is \( p(t)=\frac{2000}{41 + 75e^{-kt}} \) (logistic). Wait, the given function is \( p(t)=\frac{2000t}{41 + 75} \) which is linear, but that would have \( p'(t)=\frac{2000}{116}>0 \) constant, \( p''(t) = 0 \). But the options suggest non - linear. Likely a typo, maybe \( p(t)=\frac{2000}{41 + 75e^{-t}} \) (logistic growth). Then as \( t
ightarrow\infty \), \( e^{-t}
ightarrow0 \), so \( p(t)
ightarrow\frac{2000}{41}\approx48.78 \)? No, wait, if it's \( p(t)=\frac{2000}{41 + 75e^{-t}} \), as \( t
ightarrow\infty \), \( p(t)
ightarrow\frac{2000}{41}\approx48.78 \), so approaches a maximum. But if the function is \( p(t)=\frac{2000t}{41 + 75} \) (linear), then it grows without bound, but the options for meaning suggest growth with slowing rate, so likely the function is logistic. Assuming the function is logistic (maybe a typo in the problem), for the long - term, if \( p(t)=\frac{L}{1 + ae^{-kt}} \) (logistic), as \( t
ightarrow\infty \), \( p(t)
ightarrow L \) (maximum). But from the options, if we consider the first part (meaning) is option C, and for long - term, if the function is logistic, the population approaches a maximum. But if the function is linear (\( p(t)=\frac{2000}{116}t \)), then it grows without bound, but the meaning part's options don't match linear. So likely the function is \( p(t)=\frac{2000}{41 + 75e^{-t}} \) (logistic). So for the first part of (c), answer is C. For long - term, if it's logistic, option A: "The population of deer approaches a maximum of \(\frac{2000}{41}\approx48.78\)" (but we'll go with the option structure).

Answer:

• Meaning of answers (first part of c): C. The population is growing, but the rate of growth is slowing down
• Long - term: A. The population of deer approaches a maximum of \(\frac{2000}{41}\) (or if we consider the original function was mis - typed, but based on the options and the meaning part, this is the reasoning)