Question

anyone familiar with the logistic model with constant harvesting? need some clarification. given dN/dt =f(x) Where f(x) = rN(1-(N/K)-h. And the solutions to f(x)=0 are (k/2R)(r (+-) sqrt(r^2-(4rh/k)). Why if h>kr/4, is dN/dt<0 always?

Answer 1

@saifoo.khan I think that this will be quick for you to explain. if you dont mind.

Answer 2

I'm sorry. Can't solve this. :S

Answer 3

@zepdrix to the rescue!!

Answer 4

i doesnt necessarily need solving. basically, i derived from h* = kr/4 the discriminant, and i know that if h>kr/4 then dN/dt <0. but i cant get my head around why.

Answer 5

is it because h>kr/4 would mean the discriminant <0 thus no real solutions. And therefore dN/dt<0? Thats the only thing i can think of, but not sure if it is legit.

Answer 6

Goes over my head. Sorry. :/

Answer 7

@tcarroll010 @satellite73 @amistre64

Answer 8

Im going to type it clearer.

Answer 9

That condition falls from the radical of the solution. If that condition holds the solution to f(x)=0 is complex which is as good as not existing for the purposes of this question. That means N' never changes sign because of intermediate value theorem. So I guess what remains is to show that there exists at least one point where N' < 0

Answer 10

@Xavier so as this is a question on modelling population growth, i can say that as the soultion becomes complex, it can not exist as population must be some real number >=0.

Answer 11

also how would you go about finding the point N'<0?

Answer 12

Well what you are looking at here is the slope of that change. You want to show that this slope is never positive. To do this you show that if at one point it is negative, and never reaches zero it can't ever be positive because to go from negative to positive you need to pass through zero assuming continuity. So to show that the slope is never zero you solve dN/dt=0 and get the condition where it has no solutions (where it is complex in this case). Then all you need to show is that dN/dt was at some point negative

Answer 13

so is it possible to show that at some point it was negative without any values for the constants?