# The number of fish in a pond lake t years after restocking is given by: y=(100)/(1+2e^(-T)). When is tge population growing most rapidly? What is the population at that time?

### 2 Answers | Add Yours

`y = 100/(1+2e^(-t))`

To find an expression for the rate of ppulation growth e can differentiate wrt t.

`y = 100(1+2e^(-t))^(-1)`

`(dy)/(dt) = 100 xx 2 xx (-1) xx e^(-t) xx (-1)xx(1+2e^(-t))^(-2)`

`(dy)/(dt) = (200e^(-t))/(1+2e^(-t))^2`

Now we have new expression for rate of growth, R

`R = (200e^(-t))/(1+2e^(-t))^2`

To find the point with maximum rate, we have to differentiate R wrt t and then check for the sign of the second derivative of R.

`(dR)/(dt) = 200[(((1+2e^(-t))^2(-1)xxe^(-t))-(e^(-t)xx2xx(-1)xxe^(-t)xx2(1+2e^(-t))))/(1+2e^(-t))^4] `

This gives,

`(dR)/(dt) = 200[((-e^(-t)(1+2e^(-t)))-(-4e^(-2t)))/(1+2e^(-t))^4] `

`(dR)/(dt) = 200[(2e^(-2t)-e^(-t))/(1+2e^(-t))^4] `

`(dR)/(dt) = 200[(e^(-t)(2e^(-t)-1))/(1+2e^(-t))^4]`

For R to be maximum. `(dR)/(dt) = 0`

This gives,

`e^(-t)(2e^(-t)-1) = 0`

`e^(-t) = 0 or 2e^(-t)-1 =0`

`e^(-t) = 0 or e^(-t) =1/2`

Finding the second derivative of R, `(d^2R)/(dt^2)` would be very complex. Therefore without finding the sign of the `(d^2R)/(dt^2)` we can just check which point is the maximum.

Checking the point` e^(-t) = 0`

`t = -1`

This cannot be true. so we have to avoid this.

Therefore `e^(-t)!=0`

`e^(-t) = 1/2`

`-t = ln(1/2) = -0.6931`

`t =0.6931`

When `tlt0.6931, (dR)/(dt) gt0`

when `t gt0.6931, (dR)/(dt) lt 0`

Therefore at t=0.6931, R is a maximum.

Therefore the population grows most rapidly at

`t = 0.6931 = -ln (1/2)`

The population at that time is,

`y = 100/(1+2e^(ln(1/2)))`

`y = 100/(1+2x1/2)`

`y = 100/2`

`y = 50`

**Therefore when population grows most rapid, the populations is 50.**

Sorry, domain is 0 to 10