Problem in Picture. Consider the population model dP / dt = 2P - P2 / 50 for a s

Question

Problem in Picture.

Consider the population model dP / dt = 2P - P2 / 50 for a species of fish in a lake. Suppose it is decided that fishing will be allowed, but it is unclear how many fishing licenses should be issued. Suppose the average catch of a fisherman with a license is 3 fish per year. What is the largest number of licenses that can be issued if the fish are to have a chance to survive in the lake? Suppose the number of fishing licenses in part (a) is issued. What will happen to the fish population-that is, how does the behavior of the population depend on the initial population?

Explanation / Answer

You have dP/dt = 2P - P^2 / 50

Now the change in population should be positive

or dP / dt > 0

then 2 P - P^2 /50 > 0

or 2P > P^2 / 50

or 100P > P^2

since P cant be zero, cancel out P

then , 100 > P

or P must be less than 100.

so if there are P number of fish in the lake, then P -100 fish must be caught

no. of license to be issues = P-100 /3

You have dP/dt = 2P - P^2 / 50

or dP / (2P - P^2/50) = dt

or 50 dP / (100P - P^2) = dt

or 50 dP / { P ( P - 100) } = -dt

Now write dP as {(P - 100) - p} dP/100

so

(50/100) * {(P - 100) - P} / { P ( P - 100) } = -dt

1 / P - 1/ (P - 100) = -dt

Integrating, you get

ln P - ln (P - 100) = -t + C

or ln { P / (P - 100) } = - t + C

or P /( P - 100) = K e^-t where k is another constant

so P = K * e^ -t * P - 100 * K * e^ -t

or P ( 1 - K e^ -t) = -100 * K * e^-t

so P = 100 * K * e^-t / ( Ke^-t - 1)

This is the population of fish given as a function of time

at t = 0

Po = 100 * K / (K -1 ) = initial population = Po (say)

or 100 K = Po * K - Po

or K * (100 - Po) = Po

K = Po / (100 - Po)

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