For G(t) = 10,000/(1+24e^-1.2t) when does the population start to decrease?I know this question is looking for the maxima for the function, but I can't quite get there. Could someone show me the...

For G(t) = 10,000/(1+24e^-1.2t) when does the population start to decrease?

I know this question is looking for the maxima for the function, but I can't quite get there. Could someone show me the answer with the steps? 

2 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

First, we will find the initial population, when t= 0

==> G(0)= 10,000/(1+24)= 10,000/25= 400

Now we need to find t such that G(t) < 400

10000/(1+24e^(-1.2t)) < 400

Use reciprocal and reverse the inequality:

==> (1+24e^(-1.2t))/10,000 > 1/400

Multiply by 10,000

==> 1+24e^(-1.2t) > 25

Subtract 1.

==> 24e^(-1.2t) > 24

==> e^(-1.2t) > 1

==> -1.2t > ln 1

==>  -1.2t > 0

==> t < 0

t< 0 that means the population drcreases when t is a negative number. But time can not be negative.

Then, for all values where t>0 , the population increases. 

Then, we notice that the population does not decrease

 

 

beckden's profile pic

beckden | High School Teacher | (Level 1) Educator

Posted on

To find when a function decreases, find when the derivative is negative.

`G'(t) = ((10000)(-1))/(1+24e^(-1.2t))^2 (24)(-1.2)e^(-1.2t) = (288,000e^(-1.2t))/(1+24e^(-1.2t))`

This is always positive, so the function never decreases, G(t) gets closer and closer to 10,000 but never reaches it.  You had trouble with finding the maximum because it's maximum is at t=oo.  The minimum is at t=0 G(0) = 10,000/25 = 400.

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question