= \frac{1000}{1 + 9e^{-0.5t}}
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- P(t) = \frac{1000}{1 + 9e^{-0.5t}}
- Find the time \(t\) when the population reaches half of its carrying capacity.
- Solution:** The carrying capacity \(K\) of the logistic model is the maximum population, which is 1000 in this case. Half of the carrying capacity is \(500\). We set \(P(t) = 500\) and solve for \(t\):
- Multiply both sides by the denominator:
- 500(1 + 9e^{-0.5t}) = 1000
- Divide both sides by 500:
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