To solve this problem, we need to find the maximum value of the function $N(t)$ that represents the number of bacteria over time, where:

$N(t) = 1000 + \frac{100t}{100 + t^2}$

Here, $t$ is time in seconds.

Step-by-Step Solution

1. Identify the Function: The function $N(t) = 1000 + \frac{100t}{100 + t^2}$ models the number of bacteria over time.

2. Find the Derivative $N'(t)$: To find the maximum value, we need to calculate the first derivative $N'(t)$ and set it to zero. This will help identify any critical points.

   $N(t) = 1000 + \frac{100t}{100 + t^2}$

   Let $f(t) = \frac{100t}{100 + t^2}$. Then, the derivative $f'(t)$ can be found using the quotient rule:

   $f'(t) = \frac{(100)(100 + t^2) - (100t)(2t)}{(100 + t^2)^2}$

   Simplifying, we get:

   $f'(t) = \frac{10000 + 100t^2 - 200t^2}{(100 + t^2)^2} = \frac{10000 - 100t^2}{(100 + t^2)^2}$

   Therefore,

   $N'(t) = f'(t) = \frac{10000 - 100t^2}{(100 + t^2)^2}$

3. Set $N'(t) = 0$: To find critical points, set $N'(t) = 0$:

   $10000 - 100t^2 = 0$

   $t^2 = 100$

   $t = \pm 10$

   Since $t$ represents time in seconds, we only consider $t = 10$ seconds.

4. Evaluate $N(t)$ at $t = 10$: Substitute $t = 10$ into $N(t)$ to find the number of bacteria:

   $N(10) = 1000 + \frac{100 \cdot 10}{100 + 10^2}$

   $= 1000 + \frac{1000}{200} = 1000 + 5 = 1005$

5. Conclusion: The maximum number of bacteria is 1005.

Would you like a more detailed explanation on any specific part of this solution?

Related Questions

1. How would the answer change if the function for $N(t)$ had a different constant added to 1000?
2. What happens to $N(t)$ as $t$ approaches infinity?
3. Could we calculate the time $t$ when the bacteria population reaches a certain threshold?
4. How does the growth rate of bacteria change at different times?
5. How would we model bacterial growth if it followed a logistic rather than the given formula?

Tip

In optimization problems, always check the boundary values and any critical points to ensure you’ve found the true maximum or minimum.