Peter's chickens got the bird flu and he is rying to stop the epidemic. He knows that the function
f(t) = 300 / ( 10 + 20e^(-1.5t) )
describes the number of his chickens who are sick t weeks after the initial outbreak.

(a) How many chickens became sick when the flu epidemic has began?

I got 10/7 for the answer. does that seem right???

(b) In how many weeks will 100 chickens be sick?

(c) What is the maximum number of chickens that will become ill?

Question

Peter's chickens got the bird flu and he is rying to stop the epidemic. He knows that the function
f(t) = 300 / ( 10 + 20e^(-1.5t) )
describes the number of his chickens who are sick t weeks after the initial outbreak.

(a) How many chickens became sick when the flu epidemic has began?

I got 10/7 for the answer. does that seem right???

(b) In how many weeks will 100 chickens be sick?

(c) What is the maximum number of chickens that will become ill?

anonymous · Answer

For part (a) we just have to plug in t=0, so:
300/(10+20)=300/30=10

Part (b): Plut t=100 and solve.

Part(c): Take the derivative and set it equal to zero.

anonymous · Answer

@soccergal12 Did that help?

anonymous · Answer

yes it did, thank you

anonymous · Answer

OK, awsome! Glad to help you!

anonymous · Answer

im just unsure about the derivative; i'm not sure how to do it entirely

anonymous · Answer

and in question b, aren't i finding t, not f(100) ?

anonymous · Answer

scratch that last question .. i'm just unsure about how to do the derivative : where do i begin?

anonymous · Answer

i got ( 9000e^-1.5t ) / (10 + 20e^-1.5t )^2

anonymous · Answer

You are right, about part (b), sorry about that, we have to do this:
100=300 / ( 10 + 20e^(-1.5t) )
And solve for t

Part (c):
$$f(t) =  300( 10 + 20e^{-1.5t} )^-1$$
so the derivative is:
$$300*(-1)*(10 + 20e^{-1.5t} )^{-2}*20e^{-1.5t}(-1.5)$$

anonymous · Answer

Which is the same as:
$$\frac{9000e^{-1.5t}}{(10+20e^{1.5t})^2}$$

anonymous · Answer

So you did it great!
Any doubt?

anonymous · Answer

okay thanks! sorry i have another question about part b : to find t, i got down to e^-1.5t = -7/20 ; could you do this question and let me know if you get the same answer? also, where do i go from here: do i ln both sides?

anonymous · Answer

$$100=\frac{300}{ ( 10 + 20e^{-1.5t} )}\rightarrow10 + 20e^{-1.5t} =\frac{300}{100}$$
So then:
$$20e^{-1.5t}=20$$
$$e^{-1.5t}=1$$

anonymous · Answer

doesn't it equal 20^e −1.5t = -7 ?

anonymous · Answer

We start from here:
$$100=\frac{300}{10+20e^{-1.5t}}$$
Then we have this:
$$10+20e^{-1.5t}=\frac{300}{100}$$
So then, 
$$20e^{-1.5t}=3-10=-7$$
You are right!
Sorry about my confusion!
But then, how do we solve it?

anonymous · Answer

I think that this function never reaches 100. Take a look here: http://www.wolframalpha.com/input/?i=100+%3D+300+%2F+%28+10+%2B+20e%5E%28-1.5t%29+%29

anonymous · Answer

so would i state that the function doesn't ever reach 100?

anonymous · Answer

I think so.
You can see it on the graph.
Any other doubt?

anonymous · Answer

I have this same question, but it has been changed to: f(t) = 300 / ( 15 - 20e^(-0.07t) ) 

When I have this information though, the answer I get for part A is negative. That doesn't make sense. Can anyone help me figure out how to solve this with the changes made?

anonymous · Answer

I get a negative answer too, and my function has changed too. could anyone help us figure this out ?

dumbcow · Answer

this looks like a logistic function
it approaches an upper limit as t goes to infinity
the term "20e^(-1.5t)" goes to 0 leaving 300/10 = 30
the maximum possible sick chickens is 30
it will never be 100 so part b) is never