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Sagot :
Let's break down the problem step by step:
### Part (a): Population at Specific Times
We are given the population function:
[tex]\[ P(t) = \frac{400t}{3t^2 + 5} \][/tex]
Step 1: Find [tex]\( P(0) \)[/tex]
[tex]\[ P(0) = \frac{400 \times 0}{3 \times 0^2 + 5} = \frac{0}{5} = 0 \][/tex]
At [tex]\( t = 0 \)[/tex], the population is [tex]\( 0 \)[/tex] thousand.
Step 2: Find [tex]\( P(1) \)[/tex]
[tex]\[ P(1) = \frac{400 \times 1}{3 \times 1^2 + 5} = \frac{400}{3 + 5} = \frac{400}{8} = 50 \][/tex]
At [tex]\( t = 1 \)[/tex], the population is [tex]\( 50 \)[/tex] thousand.
Step 3: Find [tex]\( P(3) \)[/tex]
[tex]\[ P(3) = \frac{400 \times 3}{3 \times 3^2 + 5} = \frac{1200}{3 \times 9 + 5} = \frac{1200}{27 + 5} = \frac{1200}{32} = 37.5 \][/tex]
At [tex]\( t = 3 \)[/tex], the population is [tex]\( 37.5 \)[/tex] thousand.
Step 4: Find [tex]\( P(8) \)[/tex]
[tex]\[ P(8) = \frac{400 \times 8}{3 \times 8^2 + 5} = \frac{3200}{3 \times 64 + 5} = \frac{3200}{192 + 5} = \frac{3200}{197} \approx 16.24 \][/tex]
At [tex]\( t = 8 \)[/tex], the population is approximately [tex]\( 16 \)[/tex] thousand (rounded to the nearest integer).
### Part (b): Horizontal Asymptote
To find the horizontal asymptote of [tex]\( P(t) \)[/tex], we need to determine the behavior of [tex]\( P(t) \)[/tex] as [tex]\( t \)[/tex] approaches infinity.
[tex]\[ P(t) = \frac{400t}{3t^2 + 5} \][/tex]
As [tex]\( t \to \infty \)[/tex], the term [tex]\( 3t^2 \)[/tex] dominates over 5 in the denominator.
[tex]\[ \lim_{t \to \infty} \frac{400t}{3t^2 + 5} = \lim_{t \to \infty} \frac{400t}{3t^2} = \lim_{t \to \infty} \frac{400}{3t} = 0 \][/tex]
Therefore, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
As [tex]\( t \to \infty \)[/tex], [tex]\( P(t) \)[/tex] approaches [tex]\( 0 \)[/tex].
### Part (c): Interpretation of the Horizontal Asymptote
The horizontal asymptote [tex]\( y = 0 \)[/tex] indicates that as time progresses indefinitely (as [tex]\( t \to \infty \)[/tex]), the population of the resort community will approach [tex]\( 0 \)[/tex] thousand. This suggests that the population cannot sustain itself in the long term under the given conditions, and it will gradually decrease to zero over time.
### Summary of Answers:
(a) At:
- [tex]\( t = 0 \)[/tex] months, the population is [tex]\( \boxed{0} \)[/tex] thousand.
- [tex]\( t = 1 \)[/tex] month, the population is [tex]\( \boxed{50} \)[/tex] thousand.
- [tex]\( t = 3 \)[/tex] months, the population is [tex]\( \boxed{37.5} \)[/tex] thousand.
- [tex]\( t = 8 \)[/tex] months, the population is approximately [tex]\( \boxed{16} \)[/tex] thousand.
(b) The horizontal asymptote of the graph is [tex]\( \boxed{0} \)[/tex], and [tex]\( P(t) \)[/tex] approaches [tex]\( \boxed{0} \)[/tex] as [tex]\( t \to \infty \)[/tex].
(c) The horizontal asymptote signifies that in the long term, the population will approach zero, indicating that the community will not be sustainable over an extended period and will eventually diminish.
### Part (a): Population at Specific Times
We are given the population function:
[tex]\[ P(t) = \frac{400t}{3t^2 + 5} \][/tex]
Step 1: Find [tex]\( P(0) \)[/tex]
[tex]\[ P(0) = \frac{400 \times 0}{3 \times 0^2 + 5} = \frac{0}{5} = 0 \][/tex]
At [tex]\( t = 0 \)[/tex], the population is [tex]\( 0 \)[/tex] thousand.
Step 2: Find [tex]\( P(1) \)[/tex]
[tex]\[ P(1) = \frac{400 \times 1}{3 \times 1^2 + 5} = \frac{400}{3 + 5} = \frac{400}{8} = 50 \][/tex]
At [tex]\( t = 1 \)[/tex], the population is [tex]\( 50 \)[/tex] thousand.
Step 3: Find [tex]\( P(3) \)[/tex]
[tex]\[ P(3) = \frac{400 \times 3}{3 \times 3^2 + 5} = \frac{1200}{3 \times 9 + 5} = \frac{1200}{27 + 5} = \frac{1200}{32} = 37.5 \][/tex]
At [tex]\( t = 3 \)[/tex], the population is [tex]\( 37.5 \)[/tex] thousand.
Step 4: Find [tex]\( P(8) \)[/tex]
[tex]\[ P(8) = \frac{400 \times 8}{3 \times 8^2 + 5} = \frac{3200}{3 \times 64 + 5} = \frac{3200}{192 + 5} = \frac{3200}{197} \approx 16.24 \][/tex]
At [tex]\( t = 8 \)[/tex], the population is approximately [tex]\( 16 \)[/tex] thousand (rounded to the nearest integer).
### Part (b): Horizontal Asymptote
To find the horizontal asymptote of [tex]\( P(t) \)[/tex], we need to determine the behavior of [tex]\( P(t) \)[/tex] as [tex]\( t \)[/tex] approaches infinity.
[tex]\[ P(t) = \frac{400t}{3t^2 + 5} \][/tex]
As [tex]\( t \to \infty \)[/tex], the term [tex]\( 3t^2 \)[/tex] dominates over 5 in the denominator.
[tex]\[ \lim_{t \to \infty} \frac{400t}{3t^2 + 5} = \lim_{t \to \infty} \frac{400t}{3t^2} = \lim_{t \to \infty} \frac{400}{3t} = 0 \][/tex]
Therefore, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
As [tex]\( t \to \infty \)[/tex], [tex]\( P(t) \)[/tex] approaches [tex]\( 0 \)[/tex].
### Part (c): Interpretation of the Horizontal Asymptote
The horizontal asymptote [tex]\( y = 0 \)[/tex] indicates that as time progresses indefinitely (as [tex]\( t \to \infty \)[/tex]), the population of the resort community will approach [tex]\( 0 \)[/tex] thousand. This suggests that the population cannot sustain itself in the long term under the given conditions, and it will gradually decrease to zero over time.
### Summary of Answers:
(a) At:
- [tex]\( t = 0 \)[/tex] months, the population is [tex]\( \boxed{0} \)[/tex] thousand.
- [tex]\( t = 1 \)[/tex] month, the population is [tex]\( \boxed{50} \)[/tex] thousand.
- [tex]\( t = 3 \)[/tex] months, the population is [tex]\( \boxed{37.5} \)[/tex] thousand.
- [tex]\( t = 8 \)[/tex] months, the population is approximately [tex]\( \boxed{16} \)[/tex] thousand.
(b) The horizontal asymptote of the graph is [tex]\( \boxed{0} \)[/tex], and [tex]\( P(t) \)[/tex] approaches [tex]\( \boxed{0} \)[/tex] as [tex]\( t \to \infty \)[/tex].
(c) The horizontal asymptote signifies that in the long term, the population will approach zero, indicating that the community will not be sustainable over an extended period and will eventually diminish.
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