When Will the Population Be a Minimum and the Rate of Change Be a Maximum?

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Discussion Overview

The discussion revolves around determining the minimum population and the maximum rate of change of a population modeled by the function f(t) = (2t-1) / (t^2-t+0.5). Participants explore the mathematical conditions necessary to find these points, as well as the implications of the function in relation to the population dynamics of a species introduced into a refuge.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes using the first and second derivatives of the function to find the minimum population and maximum rate of change.
  • Another participant questions the notation used and suggests verifying the conditions for a differentiable function to have a minimum.
  • A participant claims to have found the minimum population at t = 1/2 and seeks clarification on how to approach finding the maximum rate of change.
  • Concerns are raised about the relationship between the function f(t) and the actual population, with one participant noting discrepancies in the values of f(0) and the maximum supported species.
  • There is a reiteration that 75 animals are introduced into the refuge, which affects the population dynamics described by f(t).

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the function and its implications for population dynamics. There is no consensus on the relationship between the function and the population, nor on the correctness of the calculations presented.

Contextual Notes

Limitations include potential misunderstandings regarding the function's definition and its application to the population model. The discussion also reflects uncertainty about the mathematical steps required to verify critical points and their nature.

huan.conchito
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Given a function f(t) = (2t-1) / (t^2-t+0.5)
max supported species = 120, and t is the time.
if 75 animals are introduced into te refuge the population changes at a rate of f(t)
What needs to be done to obtain
a ) Find when the population will be a minimum
and to find
b) When the rate of change of the population will be a maximum
 
Last edited:
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Theoretically

a)f'(x)=0,f''(x)>0
b)f''(x)=0,f'''(x)<0...

Daniel.
 
i still don't get it
 
Are you sure you don't mean [itex]f(t)[/itex]?

I'll assume that you do. In that case, what conditions are necessary for a differentiable function to have a minimum at a point? Look at your notes if you don't remember.
 
No need to look at the notes,i wrote them,before he's edited his post & spell out the function...

Daniel.
 
yup is f(t)
 
Then solve it...Set those derivatives to zero and verify the nature of the critical points.

Daniel.
 
i found the minimum at t= 1/2
how do i go about part b?
for part b i got
f''(t)= 2(2t-1)
2(2t-1)=120 -> t =30.5
is that the time of the max population?
 
Last edited:
huan.conchito said:
Given a function f(x) = (2t-1) / (t^2-t+0.5)
Okay we have a function.

max supported species = 120, 75 are initially introduced.
and t is the time.

What needs to be done to obtain
a ) Find when the population will be a minimum
and to find
b) When the rate of change of the population will be a maximum

?? What happened to f? what does f have to do with the population?
Have you left something out- like f(t) is the population at time t? But I notice that f(0)= -2, not 75 and the maximum of f is 2, not 120. What does f have to do with the population?
 
  • #10
if 75 animals are introduced into te refuge the population changes at a rate of f(t)
 

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