# optimzation problem #24

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• Aug 13th 2009, 09:23 AM
skeske1234
optimzation problem #24
After a lake has been stocked with 1000 trout, the trout population is modelled by the function N(t)=1000 + (10000t^2)/(t^2+100) where t is the time in years.

A) what is the trout population after a long period of time?
B) when is the population increasing most rapidly?
C) sketch the graph of N(t) and illustrate the results in parts A) and B)
• Aug 13th 2009, 10:58 AM
skeeter
Quote:

Originally Posted by skeske1234
After a lake has been stocked with 1000 trout, the trout population is modelled by the function N(t)=1000 + (10000t^2)/(t^2+100) where t is the time in years.

A) what is the trout population after a long period of time?
B) when is the population increasing most rapidly?
C) sketch the graph of N(t) and illustrate the results in parts A) and B)

this one is all set up for you ... do it.
• Aug 13th 2009, 11:14 AM
skeske1234
Quote:

Originally Posted by skeeter
this one is all set up for you ... do it.

starting with A).. what do they mean by long time
• Aug 13th 2009, 12:46 PM
skeeter
Quote:

Originally Posted by skeske1234
starting with A).. what do they mean by long time

what value does $\displaystyle N(t)$ approach as $\displaystyle t \to \infty$ ?
• Aug 13th 2009, 01:04 PM
skeske1234
Quote:

Originally Posted by skeeter
what value does $\displaystyle N(t)$ approach as $\displaystyle t \to \infty$ ?

N($\displaystyle \infty$) =1000+ [10000($\displaystyle \infty^2$)/($\displaystyle \infty^2$) +100]

sorry but how do i solve this?

when i found the common demoninator, i ended up getting

[11000($\displaystyle \infty^2$)+100000]/[($\displaystyle \infty^2$) +100]
• Aug 13th 2009, 01:20 PM
skeeter
have you not studied limits and end behavior of functions? do you know what a horizontal asymptote is and how to find it?

some lessons you need to review ...

Limits and Infinity

Horizontal Asymptotes
• Aug 13th 2009, 01:33 PM
skeske1234
i got 1100, the back of the book has 11000
what did you get?
• Aug 13th 2009, 01:34 PM
skeeter
Quote:

Originally Posted by skeske1234
i got 1100, the back of the book has 11000
what did you get?

back of the book is correct.
• Aug 13th 2009, 01:39 PM
skeske1234
[(1000)(100)+10000]/100
What is wrong here with my equation?

Part B)

N'(t)=[(20,000t)(t^2+100)-(20000t^3)]/[(t^2+100)^2]
=[20,000t(t^2+100-t^2)]/(t^2+100)^2]
=[20,000(100)]/[t^2+100)^2]
N'(t)=0 when t=

and with part b, what did I do wrong here? can you please check? thank you
• Aug 13th 2009, 01:56 PM
skeeter
Quote:

Originally Posted by skeske1234
After a lake has been stocked with 1000 trout, the trout population is modelled by the function N(t)=1000 + (10000t^2)/(t^2+100) where t is the time in years.

A) what is the trout population after a long period of time?
B) when is the population increasing most rapidly?
C) sketch the graph of N(t) and illustrate the results in parts A) and B)

part (b) ... the problem did not ask to find when N(t) is a maximum ...
• Aug 13th 2009, 02:26 PM
skeske1234
edit: think i got it now...almost not there. no
• Aug 13th 2009, 02:31 PM
skeeter
Quote:

Originally Posted by skeske1234
ok well i don't know what to do to answer the problem.. or how to find when the population is increasing most rapidly.

you quit too easy.

http://www.mathhelpforum.com/math-he...oblem-2-a.html
• Aug 13th 2009, 02:52 PM
skeske1234
1100=100+(10,000t^2)/(t^2+100)

my answer: t=10(sqrt10)/3

book says: t=10(sqrt3)/3

I seem to be getting the squareroot 10 instead of 3..........is there supposed to be the 3?
• Aug 13th 2009, 04:25 PM
skeeter
Quote:

Originally Posted by skeske1234
After a lake has been stocked with 1000 trout, the trout population is modelled by the function N(t)=1000 + (10000t^2)/(t^2+100) where t is the time in years.

B) when is the population increasing most rapidly?

$\displaystyle N'(t) = \frac{2000000t}{(t^2+100)^2}$

N'(t) is the rate of growth. the question wants to know when N'(t) is a maximum ...

$\displaystyle N''(t) = \frac{2000000(100-3t^2)}{(t^2+100)^3}$

$\displaystyle N''(t) = 0$ when $\displaystyle t^2 = \frac{100}{3}$

$\displaystyle t = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3}$
• Aug 13th 2009, 04:59 PM
skeske1234
Quote:

Originally Posted by skeeter
$\displaystyle N'(t) = \frac{2000000t}{(t^2+100)^2}$

N'(t) is the rate of growth. the question wants to know when N'(t) is a maximum ...

$\displaystyle N''(t) = \frac{2000000(100-3t^2)}{(t^2+100)^3}$

$\displaystyle N''(t) = 0$ when $\displaystyle t^2 = \frac{100}{3}$

$\displaystyle t = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3}$

Why do we have to take the derivative twice here to find the maximum? normally we just take the derivative (once) to find it..what is different about this case?
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