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Math Help - Pronumeral question

  1. #1
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    Pronumeral question

    Another problem, this one's been digging at me for a week or two.
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  2. #2
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    Quote Originally Posted by Naur View Post
    Another problem, this one's been digging at me for a week or two.
    Yes, it should be t^2. They have probably made a typo.
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  3. #3
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    But the answer fits with the formula, and my calculator also seems to agree.
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  4. #4
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    Quote Originally Posted by Naur View Post
    But the answer fits with the formula, and my calculator also seems to agree.
    Given the form you wrote for N'(t) it should be a t^2. Perhaps we should see the whole question.

    -Dan
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  5. #5
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    There we go. It concerns finding the maximum or minimum stationary points by finding the derivative, and putting it =0.
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  6. #6
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    Using the rules for differentiation:
    (500+100 e^{-t/12} t)'
    500'+(100 e^{-t/12} t)'
    100 \left(e^{-t/12} t\right)'
    100(t(e^{-t/12})'+e^{-t/12}t')
    100 \left(e^{-t/12} t \left(-\frac{t}{12}\right)'+e^{-t/12}\right)
    100 \left(-\frac{1}{12} e^{-t/12} t t'+e^{-t/12}\right)
    100 \left(e^{-t/12}-\frac{1}{12} e^{-t/12} t\right)
    100 e^{-t/12}-\frac{25}{3} e^{-t/12} t
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  7. #7
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    Err. Little confused.
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  8. #8
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    In my explanation I use:
    a-constant
    x,y-variables

    Your function is f(x)=500+100 e^{-t/12} t
    So f'(x)=(500+100 e^{-t/12} t)'
    f'(x)=500'+(100 e^{-t/12} t)' from (x+y)'=x'+y'
    500'=0
    f'(x)=(100 e^{-t/12} t)'
    f'(x)=100 \left(e^{-t/12} t\right)' from (ax)'=ax' (here x=e^{-t/12} t)
    f'(x)=100(t(e^{-t/12})'+e^{-t/12}t') from (xy)'=x'y+xy' (here x=e^{-t/12} and y=t)
    t'=1
    f'(x)=100 \left(e^{-t/12} t \left(-\frac{t}{12}\right)'+e^{-t/12}\right) from (e^{y})'=y'e^{y}
    f'(x)=100 \left(-\frac{1}{12} e^{-t/12} t t'+e^{-t/12}\right) from \left(\frac {x}{a}\right)'=\frac {1}{a}x'
    t'=1
    f'(x)=100 \left(e^{-t/12}-\frac{1}{12} e^{-t/12} t\right)
    f'(x)=100e^{-t/12}-\frac{100}{12} e^{-t/12} t
    f'(x)=100 e^{-t/12}-\frac{25}{3} e^{-t/12} t
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  9. #9
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    Okay, that all makes sense. That's as far as I got before the book confused me.
    Could you please give an answer for the maximum population? Like I said, though it seems the book is doing it wrong, they have an answer that seems right.

    Ps. Sorry I couldn't make it back before now, I had a friend down.
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  10. #10
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    Ok. We have:
    f(t)=500+100e^{-t/12}t
    f'(t)=e^{-t/12}-\frac{1}{12} e^{-t/12} t
    We have extremums in the points where f'(t)=0

    e^{-t/12}-\frac{1}{12} e^{-t/12} t=0
    -\frac{1}{12} e^{-t/12} (-12+t)=0

    t-12=0
    t=12

    -\frac{1}{12} e^{-t/12}=0
    e^{-t/12}=0
    In this equation there is no solution for t therefore we have t_{\max /\min }=12

    Now we just need to find if we hava a max or min at point t=12
    We check the values of f'(t) for two points. One of them smaller than t ant the other larger.

    f'(0)=1>0
    f'(24)= -\frac{1}{e^2}<0

    So the derivative at point t=0 is greater than 0, which means that the function is increasing. And at point t=24 is smaller than 0, which means the function is decreasing. In other words f(12)=max because first the function is increasing, it reaches the max and then begins to decrease.
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  11. #11
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    Okay then. So the answer is 12....I still wonder how the book got the right answer by doing it wrong, and how my answer doesn't seem to fit. I mean, I'm doing it a different way to how you've done it, and I could swear it's all right except for that squared sign.
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  12. #12
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    They have a typo. It should be (-\frac {t}{12})' instead of (-\frac {t}{12})
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  13. #13
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    I'm still at a loss as to why my working isn't providing the right answer.
    Care to take a look?
    :O! Nevermind, I just worked it out. Sorry I didn't pick it up as quickly as I would've liked, it's just that we have different ways of explaining things.
    I'll show you my page of working, it explains it all pretty well.


    Thank you very much. I shall give you thanks
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