Question
A lake is stocked with 569 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 3,556 . The population of fish in the lake after time \( t \), in months, is given by the function \( P(t)=\frac{3,556}{1+5.3 e^{-0.27 t}} \). Find the population after 1 months.
Ask by Crawford Robbins. in the United States
Mar 20,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
After 1 month, there are approximately 705 fish in the lake.
Solution
We are given the logistic population model
\[
P(t)=\frac{3556}{1+5.3e^{-0.27t}},
\]
and we need to find the population after \( t = 1 \) month. Substituting \( t = 1 \) into the model, we have
\[
P(1)=\frac{3556}{1+5.3e^{-0.27}}.
\]
**Step 1. Compute the exponential term**
Calculate
\[
e^{-0.27}.
\]
Using a calculator, we approximate:
\[
e^{-0.27}\approx 0.763.
\]
**Step 2. Multiply by 5.3**
Multiply this result by 5.3:
\[
5.3 \times 0.763 \approx 4.0439.
\]
**Step 3. Compute the denominator**
The denominator becomes
\[
1 + 4.0439 \approx 5.0439.
\]
**Step 4. Find \( P(1) \)**
Substitute the denominator back into the formula:
\[
P(1) = \frac{3556}{5.0439} \approx 705.
\]
Thus, the population after 1 month is approximately \(705\) fish.
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Simplify this solution Beyond the Answer
To find the population after 1 month, we need to substitute \( t = 1 \) into the function \( P(t) \). Starting with the formula: \[ P(t) = \frac{3,556}{1 + 5.3 e^{-0.27 t}} \] Substituting \( t = 1 \): \[ P(1) = \frac{3,556}{1 + 5.3 e^{-0.27(1)}} \] Calculating \( e^{-0.27} \) using a calculator gives approximately \( 0.763 \): \[ P(1) = \frac{3,556}{1 + 5.3 \times 0.763} \] \[ P(1) = \frac{3,556}{1 + 4.0429} \] \[ P(1) = \frac{3,556}{5.0429} \] \[ P(1) \approx 704.51 \] So, the population of fish after 1 month is approximately 705 fish (rounding to the nearest whole fish).
