Expert: Paul Klarreich - 4/7/2006

Question
how does my answer key get "5 million"?
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Followup To
Question -
Hi Paul,
A certain species of fish will grow from x million to x(15-x) million each year. In order to sustain a steady catch each year, a limit of x(15-x)-x million fish are to be caught ,leaving x million fish to reproduce each year. What is the number of fish which should be left to reproduce each year so that the maximum catch may be sustained from year to year?i couldn't really even set up an equation to solve!Please explain how to solve this problem. Thank you very much.
Answer -
Hi, Jeff,

You wrote (sort of)

A certain species of fish will grow from x to x(15-x) each year. In order to sustain a steady catch each year, a limit of x(15-x)-x  fish are to be caught, leaving x fish to reproduce each year.

What is the number of fish which should be left to reproduce each year so that the maximum catch may be sustained from year to year?

I couldn't really even set up an equation to solve!  Please explain how to solve this problem. Thank you very much.

>> I am not sure just what the problem is.  It sounds as if the problem is already answered in the question.  You catch  x(15-x)-x  fish each year.  

You are saying that

dx/dt = 15x - x^2 - x = 14x - x^2, if they are left alone.

But you want the population, as well as the 'catch' to be constant from year to year, in which case you will have:

dx/dt = 0.

So the catch rate should be  dx/dt or  15x - x^2.

I can't come up with anything more clever than that.


Answer
Hi, Jeff,
Sorry, but I can't come up with anything really good on this one.  Normally, growth of a species follows the exponential law, assuming that

x = x0 exp(kt),

where  x0 is the number of fish(?) at t = 0, and k is determined from initial and later conditions.  So you can write:

At  t = 0,  x = x0,
At  t = 1,  x = x0 exp(k) = x0(15- x0)

Then  exp k = 15 - x0 and so

k = ln(15 - x0)

So your basic growth is  x = x0 exp(t ln(15 - x0))

But that's as far as I got.