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Math Help - Optimisation Problem?

  1. #1
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    Optimisation Problem?

    Vehicle A is 25km due east of vehicle B. A moves due west at 60km/h and B moves due north at 80km/h. Find an expression for the distance of separation for the two vehicles t hours later. After how many minutes is this separation distance a minimum and what is this minimum distance?

    Im really just not sure how to find the expression, but i think i know how to find at how many minutes the separation distance is minimum (differentiate the expression then make f'(x) = 0. Right?)

    I found the expression for distance as d = 25 + 85x, but this is wrong. Apparently (according to my teacher) i need to use the Pythagoras theorem, so i guess that means im finding displacement and not distance.... now that i reread the question thats actually pretty obvious because its distance of separation not distance travelled. :/. oh well.

    Anywho, any help in finding that expression is appreciated. Thanks.
    Last edited by caitie; July 31st 2008 at 06:45 AM.
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  2. #2
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    Distance equals rate times time, right?. So, after t hours A have moved

    25-60t km and B has moved 80t km.

    The distance between them is, by ol' Pythagoras,

    D^{2}=(25-60t)^{2}+(80t)^{2}

    Now, can you finish?.
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  3. #3
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    Im unsure how to differentiate this (very new to it- just started intro-calc a couple of weeks ago) but i think im on the right track.

    s = √((25-60t) + (80t))
    s = √(625 - 3000t + 10000t)
    s = 25 - 54.77t^(1/2) + 100t
    s' = -27.58/(√t) + 100
    0 = -27.58/(√t) + 100
    100√t = 27.58
    t = 0.075 h
    t = 4.5 min


    but this is wrong, the answer is double this - its 9. Im really not sure what im not getting, this question has been bugging me all night. :/

    thankyou for your help
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  4. #4
    Eater of Worlds
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    We can differentiate with out the radical to make it easier.

    Expand (25-60t)^{2}+(80t)^{2}=1000t^{2}-3000t+625

    Differentiate: 20000t-3000=20t-3

    20t-3=0

    t=\frac{3}{20}=.15

    60(\frac{3}{20})=9

    Plug into Pythagoras to find the distance.
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