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Math Help - optimzation problem #8

  1. #1
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    optimzation problem #8

    A train leaves the station at 10:00 and travels due south at a speed of 60km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two closest together?

    totally stuck on this one.
    Last edited by mr fantastic; August 12th 2009 at 03:47 PM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    A train leaves the station at 10:00 and travels due south at a speed of 60km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two closest together?

    totally stuck on this one.
    let d_1 = distance first train is from the origin (station)

    d_1 = 60t

    d_2 = distance second train is from the same station

    d_2 = 45 - 45t


    let r = straight line distance between the two trains between 10:00 and 11:00

    r^2 = d_1^2 + d_2^2

    minimize r^2 and r will also be minimized.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    let d_1 = distance first train is from the origin (station)

    d_1 = 60t

    d_2 = distance second train is from the same station

    d_2 = 45 - 45t


    let r = straight line distance between the two trains between 10:00 and 11:00

    r^2 = d_1^2 + d_2^2

    minimize r^2 and r will also be minimized.
    Ok, can you or someone check my work below because I can't seem to get the answer with the back.

    I said..

    Let x be the distance between A and B in km
    Let a and b be the distance travelled by A and B in km

    Min x at t=?
    V=d/t
    For A
    a=60t
    For B
    b=45t

    b^2+a^2=x^2
    (45t)^2+(60t)^2=x^2
    x=(5625t^2)^0.5
    x'=5625t/[(5625t^2)^0.5]

    x'=0 when t=0

    domain of t
    10<=t<=11

    x(10)=750 km
    x(11)=825 km

    therefore the min distance between train A and train B is 750 km at 10:00

    BUT>>> the back of the book's answer is: t=0.36 h

    How do i get this and what did I do wrong?
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    Ok, can you or someone check my work below because I can't seem to get the answer with the back.

    I said..

    Let x be the distance between A and B in km
    Let a and b be the distance travelled by A and B in km

    Min x at t=?
    V=d/t
    For A
    a=60t
    For B
    b=45t

    b^2+a^2=x^2
    (45t)^2+(60t)^2=x^2 this equation is wrong
    take another look at my original post ...

    r^2 = (60t)^2 + (45 - 45t)^2<br />

    minimizing r^2 will minimize r ...

    take the derivative of the RHS and set = 0

    2(60t)(60) + 2(45-45t)(-45) = 0

    3600t - 2025 + 2025t = 0

    5625t = 2025

    t = \frac{2025}{5625} = \frac{9}{25} = .36 hr
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  5. #5
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    im still not sure with the distance of train 2 formula

    you have 45-45t =d_2
    I don't really understand this... why do we take 45 away from 45t? or the speed away from the distance?
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  6. #6
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    Quote Originally Posted by skeske1234 View Post
    im still not sure with the distance of train 2 formula

    you have 45-45t =d_2
    I don't really understand this... why do we take 45 away from 45t? or the speed away from the distance?
    the distances, d_1 and d_2 are measured from the origin.

    the first train starts at the origin and moves away from the origin at 60 km/hr

    d_1 = 60t

    the second train started 45 km away from the origin, and heads back toward the origin at 45 km/hr.

    d_2 = 45 - 45t

    you see, after 1 hr, d_2 = 45 - 45(1) = 0 km from the origin. it made it back.
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