George Bake's question at Yahoo Answers regarding the Ricker curve

[MarkFL]

Here is the question:

Calculus Word Problem?

The number of offspring in a population may not be a linear function of the number of adults. The Ricker curve, used to model fish populations, claims that y=axe^-bx , where x is the number of adults, y is the number of offspring, and ^a and ^b are positive constants.

a.) Find and classify the critical point of the Ricker curve

Here is a link to the question:

Calculus Word Problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello George Bake,

We are given the Ricker curve:

$$y=axe^{-bx}$$

To find the critical point, we need to equate the first derivative to zero:

$$y'=a\left(x\left(-be^{-bx} \right)+(1)e^{-bx} \right)=ae^{-bx}(1-bx)=0$$

Since $$0<ae^{-bx}$$ for all real $x$, the only critical value comes from:

$$1-bx=0\,\therefore\,x=\frac{1}{b}$$

Using the first derivative test, we may observe:

$$y'(0)=ae^{-b\cdot0}(1-b\cdot0)=a>0$$

$$y'\left(\frac{2}{b} \right)=ae^{-b\cdot\frac{2}{b}}(1-b\cdot\frac{2}{b})=-ae^{-2}<0$$

Hence the critical point is a global maximum, and is at:

$$\left(\frac{1}{b},y\left(\frac{1}{b} \right) \right)=\left(\frac{1}{b},\frac{a}{be} \right)$$

To George Bake and any other guests viewing this topic, I invite and encourage you to post other calculus problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.