Population Dynamics

[edwardsmit]

I'm having some trouble with a Differential Equations Population problem.

Here is the background for the question..

The DE governing a fish pop. P(t) with harvesting proportional to the population is given by:
P'(t)=(b-kP)P-hP
where b>0 is birthrate, kP is deathrate, where k>0, and h is the harvesting rate. Model assumes that the death rate per individual is proportional to the pop. size. An equilibrium point for the DE is a value of P so that P'(t)=0.

I worked out the integral to be the following:

1 / [(b-h)-kP]P dp = dt

equals:
(lnP - ln(b-h-kP)) / (b-h) + C

However, I'm having trouble answering this part of the question..
Determine h so that Y is maximized, and find this Y. This is the maximum sustainable yield.

How would I go about solving that part? Any help would be greatly appreciated.

 

[Deleted member 4993]

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http://www.mathhelpforum.com/math-help/differential-equations/96176-population-dynamics.html

I'm having some trouble with a Differential Equations Population problem.

Here is the background for the question..

The DE governing a fish pop. P(t) with harvesting proportional to the population is given by:
P'(t)=(b-kP)P-hP
where b>0 is birthrate, kP is deathrate, where k>0, and h is the harvesting rate. Model assumes that the death rate per individual is proportional to the pop. size. An equilibrium point for the DE is a value of P so that P'(t)=0.

I worked out the integral to be the following:

1 / [(b-h)-kP]P dp = dt <<< What is the difference between "P" and 'p'?
equals:
(lnP - ln(b-h-kP)) / (b-h) + C <<< What does this function equal to?

However, I'm having trouble answering this part of the question..
Determine h so that Y is maximized, and find this Y. This is the maximum sustainable yield.

What is Y? Where is Y?

How would I go about solving that part? Any help would be greatly appreciated.

 

[edwardsmit]

sorry, it should have been DP/dt.

(lnP - ln(b-h-kP)) / (b-h) + C = t i believe for this one.

But Y is the maximum sustainable yield for the fish population.

so i believe this would be... (i think you can drop that C, but i'm not positive.)

Y = (lnP - ln(b-h-kP)) / (b-h)

Any help on how I can find that?

 

[Deleted member 4993]

sorry, it should have been DP/dt.

(lnP - ln(b-h-kP)) / (b-h) + C = t i believe for this one.

But Y is the maximum sustainable yield for the fish population.

so i believe this would be... (i think you can drop that C, but i'm not positive.)

Y = (lnP - ln(b-h-kP)) / (b-h)

Any help on how I can find that?

What is the condition for maximum -( your problem states the condition -- dP/dt = 0)

 

[edwardsmit]

So you're saying I set it up as
0 = (lnP - ln(b-h-kP)) / (b-h) and solve for h? I still don't get how to find a maximum with that though..

 

[Deleted member 4993]

So you're saying I set it up as
0 = (lnP - ln(b-h-kP)) / (b-h) and solve for h? I still don't get how to find a maximum with that though..

No -- t is not set to zero - it is

dP/dt = 0 .....think a little bit. What expression did you start with?

 

[edwardsmit]

0 = (b-kP)P-hP, right?