Carrying capacity for a population is estimated at 500, the population size is currently 400, and rmax = 0.01. What is d0N`//`dt?

Options

A. 0.01
B. 0.8
C. 8
D. 50

Answer

To solve the problem, we need to calculate the rate of change of the population size (dN/dt) using the logistic growth model. The formula for dN/dt is given by:

\[ \frac{dN}{dt} = r \cdot N \cdot \left( \frac{K - N}{K} \right) \]

Where:
- \( r \) = intrinsic growth rate (rmax)
- \( N \) = current population size
- \( K \) = carrying capacity

### Step-by-Step Solution:

1. **Identify the given values**:
- Carrying capacity \( K = 500 \)
- Current population size \( N = 400 \)
- Maximum intrinsic growth rate \( r = 0.01 \)

2. **Substitute the values into the formula**:
\[ \frac{dN}{dt} = r \cdot N \cdot \left( \frac{K - N}{K} \right) \]
Plugging in the values:
\[ \frac{dN}{dt} = 0.01 \cdot 400 \cdot \left( \frac{500 - 400}{500} \right) \]

3. **Calculate \( K - N \)**:
\[ K - N = 500 - 400 = 100 \]

4. **Calculate \( \frac{K - N}{K} \)**:
\[ \frac{K - N}{K} = \frac{100}{500} = 0.2 \]

5. **Now substitute this value back into the equation**:
\[ \frac{dN}{dt} = 0.01 \cdot 400 \cdot 0.2 \]

6. **Perform the multiplication**:
\[ \frac{dN}{dt} = 0.01 \cdot 400 = 4 \]
\[ \frac{dN}{dt} = 4 \cdot 0.2 = 0.8 \]

7. **Final answer**:
\[ \frac{dN}{dt} = 0.8 \]

### Conclusion:
The rate of change of the population size \( \frac{dN}{dt} \) is 0.8.