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Math Help - max/min problem #3

  1. #1
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    max/min problem #3

    A flu epidemic breaks out in a city. The fraction of the population that is infected at time t, in weeks, is given by the function f(t)=64t/(8+t)^3. What is the largest fraction of the population that is infected during the first 10 weeks? Assume that t=0 when the epidemic starts.

    Ok.. I think I have my method correct for this, but I am not exactly sure, so can someone verify my method and calculations to see if I have the correct answer? Thank you very much in advance


    M_t=f'(t)=[64(8+t)^3-(64t)(3(8+t)^2)(1)]/((8+t)^3)^2
    =[64(t^3+21t^2+144t+320)]/(8+t)^6

    Critical Numbers
    f'(t)=0 when t = -5 or t = -8
    f'(t)=undefined when t = -8

    Reject answers above because all negative.. time cannot be negative
    In this case, domain is 0<=t<=10.. therefore

    Test end points
    f(0)=0
    f(10)=640/5832=0.1097

    Therefore the largest fraction of the population that is infected during the first ten weeks is 0.1097 when it is the tenth week.
    Last edited by mr fantastic; August 12th 2009 at 03:55 PM. Reason: Changed post title
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  2. #2
    Super Member Matt Westwood's Avatar
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    Don't know where you get your working after "Critical numbers". I make it the same result after differentiating, but all you need to do is note that when f'(t)=0 you're just interested in the top, and 64(8+t)^2 is a common factor of both terms. You're left with:
    8+t = 3t
    (unless I've made a mistake)
    and it should be fairly straightforward from there.
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  3. #3
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    Quote Originally Posted by Matt Westwood View Post
    Don't know where you get your working after "Critical numbers". I make it the same result after differentiating, but all you need to do is note that when f'(t)=0 you're just interested in the top, and 64(8+t)^2 is a common factor of both terms. You're left with:
    8+t = 3t
    (unless I've made a mistake)
    and it should be fairly straightforward from there.

    when you say.. "8+t=3t" do you mean...
    =[64(8+2)^2][8+t-3t]
    and if i continue do you get
    =[-128(t-4)]/[(8+4)^4]

    and then

    Critical Numbers
    f'(t)=0 when t=4

    Test
    f(4)=256/1728=0.15
    f(0)=0
    f(10)=640/5852=0.11

    Therefore the largest fraction of the population that is infected during the first 10 weeks is 0.15 people when it is the fourth week.


    OK. is my final answer correct above? does it now match yours?
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  4. #4
    Super Member Matt Westwood's Avatar
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    You're making it too complicated.

    You've got:

    f(t)= \frac {64t} {(8+t)^3}

    First thing you want to do is find t when f'(t) = 0. Ignore everything else till you found this.

    Diffing it:
    f'(t)= \frac{64(8+t)^3 - (64t)(3(8+t)^2)} {(8+t)^6}

    So far so good. DON'T multiply it out, but equate it to 0 and see what you get:

    \frac{64(8+t)^3 - (64t)(3(8+t)^2)} {(8+t)^6} = 0

    Assume t \ne -8 so you can multiply by (8+t)^6 and you got:

    64(8+t)^3 = (64t)(3(8+t)^2)

    Divide both sides by 64(8+t)^2 and what you got left is:
    8+t = 3t

    so no need for all that messy mucking about with big numbers (big: 3 digits or more).

    This indeed gives you t = 4 which you can plug into your initial formula for f(t)= \frac {64t} {(8+t)^3} and see what you get (eugh, nasty big numbers).

    The moral of the story is: don't multiply out all your neatly factored terms until you really need to. Cancel out whatever you can as soon as you get the opportunity. It often makes it so much simpler.

    Hope this helps.
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