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Thread: Shortest possible diatnace to link these four towns

  1. #1
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    Shortest possible diatnace to link these four towns

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  2. #2
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    Re: Shortest possible diatnace to link these four towns

    The lowest I've been able to get is 652.93 miles and a final answer of 19588 million £
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  3. #3
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    Re: Shortest possible diatnace to link these four towns

    I get a substantially higher minimum figure. I don't see how you can possibly get an answer that is not a integer number of miles - please explain.
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    Re: Shortest possible diatnace to link these four towns

    I linked the towns in this kind of fashion

    A.`````````````````````B
    \ ____________ /
    /``````````````````` \
    B```````````````````````C
    Ignore the `

    Then I found the minimum if this function :

    4(√(106.5^2+n^2))+(284-2n), 0≤n≤142
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  5. #5
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    Re: Shortest possible diatnace to link these four towns

    I get 21600 million dollars.

    355^2 = 213^2 + 284^2
    ABCD (labelled by beginning letters of each town) is a rectangle.

    The shortest road would be a cross through the rectangle.. so 355 + 355 = 710 miles
    OR

    AD (213 miles) + DC (284 miles) + CB (213 miles)
    or a similar variation
    DA (213 miles) + AB (284 miles) + BC (213 miles)

    In any case.. the combined length of roads (minimum) seems to be 710 miles.

    I donít see how it could be less than that.
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  6. #6
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    Re: Shortest possible diatnace to link these four towns

    road miles, $m$ ...

    $m = 4\sqrt{x^2+106.5^2} + (284-2x)$

    $\dfrac{dm}{dx} = \dfrac{4x}{\sqrt{x^2+106.5^2}} - 2$

    $\dfrac{dm}{dx} = 0 \implies \dfrac{2x}{\sqrt{x^2+106.5^2}} = 1$

    $4x^2 = x^2+106.5^2$

    $x = \dfrac{106.5}{\sqrt{3}} \implies m \approx 653$ miles
    Attached Thumbnails Attached Thumbnails Shortest possible diatnace to link these four towns-road_cost.png  
    Thanks from metlx
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    Re: Shortest possible diatnace to link these four towns

    thanks for your guide...... its the best explain
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