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Math Help - optimzation problem #24

  1. #1
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    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)
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    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.
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    Quote Originally Posted by skeeter View Post
    this one is all set up for you ... do it.
    starting with A).. what do they mean by long time
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    Quote Originally Posted by skeske1234 View Post
    starting with A).. what do they mean by long time
    what value does N(t) approach as t \to \infty ?
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    Quote Originally Posted by skeeter View Post
    what value does N(t) approach as t \to \infty ?
    N( \infty) =1000+ [10000( \infty^2)/( \infty^2) +100]

    sorry but how do i solve this?

    when i found the common demoninator, i ended up getting

    [11000( \infty^2)+100000]/[( \infty^2) +100]
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    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
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    i got 1100, the back of the book has 11000
    what did you get?
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  8. #8
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    Quote Originally Posted by skeske1234 View Post
    i got 1100, the back of the book has 11000
    what did you get?
    back of the book is correct.
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    [(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
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  10. #10
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    Quote Originally Posted by skeske1234 View Post
    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 ...
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    edit: think i got it now...almost not there. no
    Last edited by skeske1234; August 13th 2009 at 03:56 PM.
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    Quote Originally Posted by skeske1234 View Post
    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
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  13. #13
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    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?
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  14. #14
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    Quote Originally Posted by skeske1234 View Post
    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?
    N'(t) = \frac{2000000t}{(t^2+100)^2}<br />

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

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

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

    t = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3}
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  15. #15
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    Quote Originally Posted by skeeter View Post
    N'(t) = \frac{2000000t}{(t^2+100)^2}<br />

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

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

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

    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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