There is a fish species that grows according to logistic growth:

where St is the number of fish in period t, S_t+1 is the number of fish in the next period (t + 1), K is natural carrying capacity (maximum fish population), and g is the intrinsic growth rate. Assume that g = 1 and K = 500. Assume that the price of fish in the market is $10 per fish. For simplicity assume that it costs $5 per fish to harvest fish.
Assume the discount rate is 10%.

a. If in period t there are 200 fish, how many fish will there be in the next period, t + 1? What is the growth of the fish population where growth equals S_t+1 − S_t?

b. Plot the growth of the fish population as a function of current fish population by solving for S_t+1 − S_t for values of S_t = 0, 50, 100, 150, 200, 250, 300, 350, 400, 450, and 500.

c. Using your answer to (b), at what level of population does growth reach its highest point? Note that this is the maximum sustained yield population level.
How much growth is there at this level of population? Note that this is the maximum sustained yield. How much profit would harvesters make if they harvested the maximum sustained yield?

d. What happens to growth if the population is at carrying capacity (S_t = K)?
What happens to growth if, for some reason, the population temporarily exceeded carrying capacity?

e. For logistic growth, the Marginal Growth Function is MG(S_t) = g ∗ (1 − 2S_t/K) Use the golden rule of growth to find the level of stock that maximizes the present value of harvest. How does this level of stock compare to the maximum sustained yield level of stock? How does the level of harvest compare?

f. Now suppose that the discount rate is zero. Use the golden rule of growth to find the level of stock that maximizes the present value of harvest. How does this level of stock compare to the maximum sustained yield level of stock?

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S+1 = 8 +8+5 + (1) * *