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For a fish population modeled by a depensation growth model with harvesting, we have

$\dfrac {dN} {dt} = F (N) - H(N)$

where

$F(N) = rN \left (\dfrac N {N_c} - 1 \right)\left( 1 - \dfrac N K \right)$

models the growth rate of the fish population without harvesting and

$H(N) = qEN$

is the rate at which fish are harvested.

I'm trying to find the sustained yield $H(N^*_3)$ and the unsustainable yield $H(N^*_2)$ as functions of effort ($E$). ($N^*_3$ is the nontrivial stable equilibrium of $N$ and $N^*_2$ is the unstable equilibrium.) I'm asking you for help.

I'm having a little trouble getting those functions. I have to determine the maximum effort and look into trends (can you recover from going above $E_\max$, etc) as well, but I don't think that'll be hard if I can figure out the equations.

Thanks for your help!

(Note TeX code: E_max changed to E_\max)

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    This article said $E_max$. I changed it to $E_\max$ by adding one character. But edits have to be at least six characters. I guess informing TeX-newbies about things like this is too trivial a thing to be worth allowing.2011-07-22
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    I didn't know you could use the slash for that. I usually would do E_{max}.2011-07-22
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    E_\max works because \max is the name of an operator which, conveniently, is typeset by LaTeX as "max" in roman type. If you wanted to have, say, $E_\mathrm{foobar}$, you could use E_\mathrm{foobar}.2011-07-22

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The equilibrium values are simply the solutions of $F(N) - H(N) = 0$, which (after dividing by $N$) is a quadratic equation.

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    I'm still confused. Aren't there just too many unknown variables?2011-07-21
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    My current thought is that we define $N = N_c$ for unstable equilibrium and $N=K$ for nontrivial stable equilibrium. then our functions are just $qEN_c$ and $qEK$. If this is the case, then wouldn't we have two straight lines?2011-07-21
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    No, the variable is $N$. The other quantities are parameters, and the solution will depend on them.2011-07-21