When there is no fishing, the growth of a population is governed by: dy/dt=2.5y(

Question

When there is no fishing, the growth of a population is governed by:

dy/dt=2.5y(1-y) where y is a fraction o the maximal number (y in [0,1]0 at time t

when fish are removed by fisherman at a rate of h proportional to the fraction of fish per year:

dy/dt=2.5y(1-y)-hy

solve for equilibrium values, and determine values for h so that the nonzero equilibrium value in (1) is larger than 0

I have those answers, which are y=0 y=1, and then y=1-h/2.5, and 2.5>h

but then how do i determine the conditions of the stability points, and once i do that, how do i know when the nonero equilibrium is stable and when extinction is stable? & then how do i calculate maximum sustainable fish harvest using optimization maximizing z=hy* where y is the expression for the nonzero equilibrium value.

Explanation / Answer

The equilibrium values of an autonomous differential equation are those values of the dependent variable for which its time derivative is equal to zero. That is, if dy/dt = f(y), then the equilibrium values of y are those for which f(y) = 0.

For your first equation, we have:

dy/dt = 2.5*y*(1-y)

Setting this equal to zero:

0 = 2.5*y*(1-y)

By inspection, this equation is satisfied when y = and when y = 1. These are the two equilibrium values of this equation.

For the next equation, we have:

dy/dt = 2.5*y*(1-y) - h*y

Again setting the derivative equal to zero and solving for y:

0 = y*[2.5*(1-y) - h]

Obviously, y = 0 is again an equilibrium solution. Now simplify to find the other solution:

0 = 2.5*(1-y) - h

h/2.5 = 1-y

y = 1 - 2h/5

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