# Application of differentiation

1. Apr 6, 2005

### huan.conchito

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: Apr 7, 2005
2. Apr 6, 2005

### dextercioby

Theoretically

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

Daniel.

3. Apr 7, 2005

### huan.conchito

i still dont get it

4. Apr 7, 2005

### Data

Are you sure you don't mean $f(t)$?

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.

5. Apr 7, 2005

### dextercioby

No need to look at the notes,i wrote them,before he's edited his post & spell out the function...

Daniel.

6. Apr 7, 2005

yup is f(t)

7. Apr 7, 2005

### dextercioby

Then solve it...Set those derivatives to zero and verify the nature of the critical points.

Daniel.

8. Apr 7, 2005

### huan.conchito

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: Apr 7, 2005
9. Apr 7, 2005

### HallsofIvy

Staff Emeritus
Okay we have a function.

?? 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. Apr 7, 2005

### huan.conchito

if 75 animals are introduced into te refuge the population changes at a rate of f(t)

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