Hey tylersmith7690.
If P - P^2/4 - EP = 0 then P = (-1 +- SQRT(1 - 4*1/4*E))/2*(-1/4)
Expanding out your version gives
(P+E)(P-1/4) = P^2 - 1/4*P + EP - 1/4*E != P - P^2/4 - EP
Double check your answer and see which solution is the right one.
The simplest useful model for fisher comes from the logistic model for population growth, together with a harvest h which is proportional to the current population P, that is,
h=EP,
where the constant E is called the effort. E measures the fraction of the population harvested, so that 0 <= E <= 1. This gives the model
dP/dt = kP(1-(P/a) - EP,
where P(t) is the number of these fish at time t year and k (the natural growth rate) and a( the carrying capacity) are constants for a particular fish polulation. In what follows take k= 1 and a= 4, for simplicity.
a) determine the equilibrium solutions for a given effort E.
my attempt.
dP/dt = P - P^2/4 - EP
dP/dt = ( P + E) (P -1/4)
equilibrium solutions are when dP/dt=0
so P(0)= -E, 1/4.
Hey tylersmith7690.
If P - P^2/4 - EP = 0 then P = (-1 +- SQRT(1 - 4*1/4*E))/2*(-1/4)
Expanding out your version gives
(P+E)(P-1/4) = P^2 - 1/4*P + EP - 1/4*E != P - P^2/4 - EP
Double check your answer and see which solution is the right one.