Answered

Use the logistic growth model [tex]f(x)=\frac{220}{1+8 e^{-2 x}}[/tex].

Find the carrying capacity.

[tex]\[\square\][/tex]



Answer :

To determine the carrying capacity in the logistic growth model given by [tex]\( f(x) = \frac{220}{1 + 8 e^{-2x}} \)[/tex], we need to understand the concept of carrying capacity in such a model.

1. Definition: The carrying capacity in a logistic growth model is the maximum population size of the species that the environment can sustain indefinitely given the food, habitat, water, and other necessary resources available in the environment.

2. Key Observation: The carrying capacity is observed when the population [tex]\( x \)[/tex] tends to infinity. Essentially, it's the stable maximum value that the function [tex]\( f(x) \)[/tex] approaches as [tex]\( x \)[/tex] increases.

3. Logistic Model Specifics:
- In the function [tex]\( f(x) = \frac{220}{1 + 8 e^{-2x}} \)[/tex], notice the numerator [tex]\( 220 \)[/tex].
- As [tex]\( x \)[/tex] tends to infinity, the term [tex]\( 8 e^{-2x} \)[/tex] in the denominator approaches zero because [tex]\( e^{-2x} \)[/tex] approaches zero.

4. Simplification as [tex]\( x \to \infty \)[/tex]:
- When [tex]\( x \to \infty \)[/tex], [tex]\( e^{-2x} \to 0 \)[/tex].
- This makes the denominator [tex]\( 1 + 8 e^{-2x} \)[/tex] approach [tex]\( 1 \)[/tex].

5. Final Carrying Capacity Calculation:
- Thus, [tex]\( f(x) \)[/tex] approaches [tex]\( \frac{220}{1} = 220 \)[/tex] as [tex]\( x \to \infty \)[/tex].

Therefore, the carrying capacity of the logistic growth model [tex]\( f(x) = \frac{220}{1 + 8 e^{-2x}} \)[/tex] is [tex]\(\boxed{220}\)[/tex].