Questions: Part 3 of 7 0 of 1 Point The number of fish in a commercial pond can be modeled by the following function F(t) = 401 - 401 e^(-0.29 t), where F(t) is the number of fish in month t. (a) Graph the function. (b) Estimate the number of fish in the pond in month 0, 1, 5, 10, and 15. (c) Is there a maximum number of fish this pond can accommodate? If yes, how many?

Solution Steps

To graph the function \( F(t) = 401 - 401 e^{-0.29 t} \), we need to set up the appropriate parameters for the graph function.

To estimate the number of fish, we substitute \( t = 0, 1, 5, 10, \) and \( 15 \) into the function \( F(t) \).

\[ F(0) = 401 - 401 e^{-0.29 \cdot 0} = 401 - 401 \cdot 1 = 0 \]

\[ F(1) = 401 - 401 e^{-0.29 \cdot 1} = 401 - 401 e^{-0.29} \approx 401 - 401 \cdot 0.7483 \approx 101.68 \]

\[ F(5) = 401 - 401 e^{-0.29 \cdot 5} = 401 - 401 e^{-1.45} \approx 401 - 401 \cdot 0.2346 \approx 307.9 \]

\[ F(10) = 401 - 401 e^{-0.29 \cdot 10} = 401 - 401 e^{-2.9} \approx 401 - 401 \cdot 0.0550 \approx 378.9 \]

\[ F(15) = 401 - 401 e^{-0.29 \cdot 15} = 401 - 401 e^{-4.35} \approx 401 - 401 \cdot 0.0129 \approx 396.8 \]

To determine if there is a maximum number of fish, we analyze the function \( F(t) = 401 - 401 e^{-0.29 t} \). As \( t \) approaches infinity, \( e^{-0.29 t} \) approaches 0, making \( F(t) \) approach 401. Therefore, the maximum number of fish the pond can accommodate is 401.

Final Answer