Solution:
Given that the number of bacteria in a culture, N, at time t is expressed as
The rate of change is expressed by taking the derivative of the above relation with respect to t.
This gives:
![N^(\prime)(t)=(dN)/(dt)=2000\left(-\frac{te^{-(t)/(20)}}{20}+e^{-(t)/(20)}\right)](https://img.qammunity.org/2023/formulas/mathematics/college/lshbcte8644oe54ntfd2kgewgr4lvelo9l.png)
Thus, when the rate of change equals zero, we have
![\begin{gathered} N^(\prime)(t)=0 \\ \Rightarrow2000\left(-\frac{te^{-(t)/(20)}}{20}+e^{-(t)/(20)}\right)=0 \\ thus\text{ we have} \\ -\frac{te^{-(t)/(20)}}{20}=-e^{-(t)/(20)} \\ multiply\text{ both sides by 20} \\ -te^{-(t)/(20)}=-20e^{-(t)/(20)} \\ cancel\text{ out similar factors,} \\ \therefore \\ t=20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d0cnf9jhmop7b1smu6zteantnhc6wquxpk.png)
Hence, the rate of change of of the number of bacteria equals zero when
![t=20](https://img.qammunity.org/2023/formulas/physics/college/f40q2f1ehpjz3n42e2sdndgpnm5ep8c8nl.png)