Chapter 5.1, Problem 19E

More on the Pacific Sardine This is a continuation of Example 5.1. In this exercise, we explore the Pacific sardine population further, using the model in Example 5.1.

a. If the current level of the Pacific sardine population is $50,000$ tons, how long will it take for the population to recover to the optimum growth level of $1.2\text{million}\text{tons}$? (Suggestion: One way to solve this is to make a new logistic formula using $K-2.4$, $r-0.338$, and $N\left(0\right)-0.05$.)

b. The value of $r$ used in Example 5.1 ignores the effects of fishing. If fishing mortality is taken into account, then r drops to $0.215$ per year (with the carrying capacity still at $2.4\text{million}\text{tons}$). Answer the question in part a using this lower value of $r$.

(Note: The population estimate of $50,000$ tons and the adjusted value of $r$ are given in the paper by Murphy (see footnote $3$ on page $347$). Murphy points out that factoring in the growth of the competing anchovy population makes the recovery times even longer, and he adds. "It is disconcerting to realize how slowly the population will recover to its level of maximum productivity ... even if fishing stops.")

Studies to fit a logistic model to the Pacific sardine population have yielded.

$N=\frac{24}{1+239e^{-0.338t}}$

where $t$ is measured in years and $N$ is measured in millions of tons of fish.

Part 1 What is $r$ for the Pacific sardine?

Part 2 According to the logistic model, in the absence of limiting factors, what would be the annual percentage growth rate for the Pacific sardine?

Part 3 What is the environmental carrying capacity $K$?

Part 4 What is the optimum yield level?

Part 5 Make a graph of $N$ versus $t$.

Part 6 At what time $t$ should the population he harvested?

Part 7 What portion of the graph is concave up? What portion is concave down?