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Question: A lake is stocked with 100 fish.

A lake is stocked with 100 fish. Let f (t) be the number of fish after t months, and suppose that y = f (t) satisfies the differential equation y’ = .0004y(1000 - y). Figure 7 shows the graph of the solution to this differential equation. The graph is asymptotic to the line y = 1000, the maximum number of fish that the lake can support. How fast is the fish population growing when it reaches one-half of its maximum population? Figure 7:
A lake is stocked with 100 fish. Let f (t) be the number of fish after t months, and suppose that y = f (t) satisfies the differential equation y’ = .0004y(1000 - y). Figure 7 shows the graph of the solution to this differential equation. The graph is asymptotic to the line y = 1000, the maximum number of fish that the lake can support. How fast is the fish population growing when it reaches one-half of its maximum population?

Figure 7:





Transcribed Image Text:

1000 800 600 400 200 y 5 10 + 15 t


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2.99

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