Math Problem Statement
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Solution
To solve this problem, let's analyze and understand the given information.
We are provided with a formula describing the number of bacteria over time :
where:
- is time in seconds,
- represents the number of bacteria at time .
Task
The question asks for the maximum number of bacteria after the initial seeding of 1000 bacteria.
Solution Approach
To find the maximum value of , we need to:
- Take the derivative of with respect to to find .
- Set to find critical points, which are potential maximum points.
- Evaluate at the critical points to find the maximum number of bacteria.
Step 1: Derivative of
The function can be differentiated using the quotient rule for the term .
- Differentiate to find .
- Solve for when .
Step 2: Find the Maximum
After finding the critical points, substitute them back into to determine the maximum number of bacteria.
Let me calculate this for you.The critical points for are and . However, since time represents seconds after the experiment starts, we only consider positive values of .
Evaluating at :
Thus, the maximum number of bacteria after the initial seeding is 1005 bacteria.
Would you like further details on the steps taken, or have any questions?
Here are five related questions to deepen understanding:
- How would the maximum number change if the initial number of bacteria was different?
- How does the rate of growth of the bacteria change as time progresses?
- What does the second derivative of tell us about the concavity and nature of the critical point?
- How would the solution change if was measured in minutes instead of seconds?
- What is the significance of the function's asymptotic behavior as ?
Tip: For any function describing growth, finding critical points and evaluating limits at large values helps understand long-term trends.
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Math Problem Analysis
Mathematical Concepts
Calculus
Optimization
Derivatives
Formulas
N(t) = 1000 + (100t) / (100 + t^2)
Theorems
First Derivative Test for Maximum
Suitable Grade Level
Grades 11-12
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