# The population on an island is represented by the function p(t) = 5000/1+ 12e^(-2t/3) t=time of year, determine the upper limit on # of deer The population on an island is represented by the...

The population on an island is represented by the function p(t) = 5000/1+ 12e^(-2t/3) t=time of year, determine the upper limit on # of deer

The population on an island is represented by the function

p(t) = 5000/1+ 12e^(-2t/3)

t=time of year, determine the upper limit on # of deer

a) t=time of year, determine the upper limit on the number of deer that the island could support. explain your reasoning

b) explain how to determinewhen will there be 3600 deer on the island. Do not solve!

3600= 5000/1+ 12e^(-2t/3)

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Given `p(t)=5000/(1+e^((-2t)/3))` :

This is a logisitics function in the form `y=a/(1+e^(-kt))` . In this form, 0 is a lower bound for the function and a is an upper bound. The function approaches zero as t tends to negative infinity,and approaches a asymptotically as t tends to positive infinity. a is the upper bound because as t increases, `e^(-kt)->0` , so the denominator approaches 1. (Note that at time t=0, there are 2500 deer as the denominator is 1+`e^0` .)

**Thus for the given function, the maximum population approaches 5000.**

As for solving for a particular population, your formulation of

`3600=5000/(1+e^((-2t)/3))` is correct. You would cross multiply, then subtract 1, then you would take the natural logarithm of both sides to eliminate the exponential, finally dividing by -2/3 to get t.

`1+e^((-2t)/3)=5000/3600`

`e^((-2t)/3)=25/18-1`

`ln[e^((-2t)/3)]=ln(7/18)`

`(-2t)/3=ln(7/18)`

`t=-3/2ln(7/18)~~1.42`