Course:MATH102/Question Challenge/2001 December Q5

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Question

In Fish River, the number of salmon (in thousands), $x$, in a given year, is linked to the number of salmon (in thousands), $y$ in the following year by the function

$\displaystyle y=Axe^{-bx}$

where $A,b$ are positive constants.

(a) For what number of salmon is there no change in the number from one year to the next?

(b) Find the number of salmon that would yield the largest number of salmon in the following year.

Hints

Hint 1 In the case that there is no change, what can you assume about $x$ and $y$?

Hint 2 The number of salmon in the following year is $y$. You know that $y$ is a function of $x$. What should you do to find the largest value of $y$?

Solution

Solution to part (a) To find the number of salmon that lead to no change the following year, set $x=Axe^{-bx}$ and solve for $x$. One solution is $x=0$ but to get others, divide through by $x$. Now we have $1=Ae^{-bx}$. ...

$ln1=lnA+ln(e^{-bx})$

$0=lnA-bx$

$bx=lnA$

$x=lnA/b$

The carrying capacity of the salmon in Fish River is $lnA/b$ in thousands of salmon, or $1000*lnA/b$ salmon.

Solution to part (b)

To find the largest value of $y$ we will take the first derivative:

${\frac {dy}{dx}}=Axe^{-bx}+Ae^{-bx}$

To find the value of $x$ which grants the maximal value of $y$ we set the derivative equal to zero and isolate for $x$:

${\frac {dy}{dx}}=A(x(-be^{-bx})+e^{-bx})=Ae^{-bx}(-bx+1)=0$

$0=-bx+1$

$x=1/b$

We know that this value produces a maximum because $-bx+1$ will be positive for values of $x$ less than $1/b$ and negative for values of $x$ greater then $1/b$.

The value of $x$ which leads to the largest value of $y$ is given by $x=1/b$.