- Represent the population growth as an exponential function \(n(y) = a \cdot b^y\).
- Determine the constants using the provided data points, yielding \(b \approx 1.0909\) and \(a \approx 50.4167\).
- Solve the equation \(600 = 50.4167 \cdot 1.0909^y\) for \(y\) to find the time at which the population hits the carrying capacity.
- Calculate \(y \approx 28.47\), which aligns with the provided option of \(\boxed{29}\).
Explanation
- Problem Analysis and Model Selection
To solve for the time \(y\) when the population reaches \(600\) organisms, we assume an exponential growth model of the form \(n(y) = a \cdot b^y\). We are given the data points: \((1, 55)\), \((2, 60)\), \((3, 67)\), and \((4, 75)\). Our objective is to determine the constants \(a\) and \(b\) that best fit the data and then find the year \(y\) for which \(n(y) = 600\).
- Determining the Model Constants
Using the first two data points, \((1, 55)\) and \((2, 60)\), we can determine the growth rate \(b\) and the initial value \(a\):
\[n(1) = a \cdot b^1 = 55\]
\[n(2) = a \cdot b^2 = 60\]
Dividing the second equation by the first, we find the growth factor:
\[b = \frac{60}{55} = \frac{12}{11} \approx 1.0909\]
Substituting \(b\) back into the first equation to solve for \(a\):
\[a \cdot \frac{12}{11} = 55 \Rightarrow a = \frac{55 \times 11}{12} = \frac{605}{12} \approx 50.4167\]
The exponential model is approximately \(n(y) = 50.4167 \cdot (1.0909)^y\).
- Calculating the Time to Reach Carrying Capacity
Now we set the population \(n(y)\) to \(600\) and solve for \(y\):
\[600 = 50.4167 \cdot (1.0909)^y\]
\[(1.0909)^y = \frac{600}{50.4167} \approx 11.9008\]
Taking the natural logarithm of both sides:
\[y \cdot \ln(1.0909) = \ln(11.9008)\]
\[y = \frac{\ln(11.9008)}{\ln(1.0909)} \approx \frac{2.4766}{0.0870} \approx 28.47\]
Given that the population growth trend in the table is slightly accelerating, the population will reach \(600\) organisms at approximately \(28.47\) years. Comparing this to the given options (61, 12, 29), the value \(29\) is the most appropriate answer.
Examples
Exponential growth models are used in many fields, such as biology to predict the spread of diseases, finance to model compound interest, and environmental science to study how quickly a population might exceed its resources. In our case, the model helps identify when a population will reach a critical carrying capacity, allowing resource managers to plan conservation efforts or adjust management strategies effectively before resources are depleted. Understanding these curves allows for informed decision-making in real-world resource allocation scenarios.