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Proof for Max/Min Problem

  1. Mar 8, 2006 #1
    We had to do a question in my Advanced Mathematics class, and the way that I did the problem was supposedly right. My teacher even did it that way. However, the answer was wrong, so my teacher showed us the correct way to do the problem.

    The dilemma I have is that BOTH ways should work, but they don't. :surprised Therefore, I am going to give both solutions to the problem, and I was wondering if you could tell me why the first one does not work.

    Question: The coordinates of points P and Q are (1,2) and (2,-3), respectively, and R is a point on the line x=-1. Find the coordinates of R so that PR + RQ is a minimum?

    Here is the graph/image I used for this problem:
    [​IMG]

    MY SOLUTION:

    Equations: PR^2 = 2^2 + y^2
    QR^2 = 3^2 + (5 - y)^2

    PR^2 + QR^2 = minimum or m
    2^2 + y^2 + 3^2 + (5 -y)(5 - y) = minimum
    4 + y^2 + 9 + 25 - 10y + y^2 = minimum
    2y^2 - 10y + 38 = minimum

    y = -b/2a = 10/4 = 5/2

    Y-coordinate of R = 2 - 5/2 = -1/2

    Therefore, the coordinates of R are (-1, -1/2)

    CORRECT SOLUTION:

    Let m = minimum

    mP'R = [ (2 - y) / (-3 + 1) ] = [ (2 - y) / -2 ]

    mRQ = [ (y + 3) / (-1 -2) ] = [ (y + 3) / -3 ]

    Note: mP'R = mRQ
    [ (2 - y) / -2 ] = [ (y + 3) / -3 ]
    = -2y - 6 = 3y - 6
    = -5y = 0
    y = 0

    The coordinates of R are therefore (-1, 0).

    *So...I have presented both solutions, and I would highly appreciate your help. I am really interested in trying to understand this dilemma.

    Thanks
     
  2. jcsd
  3. Mar 9, 2006 #2

    NateTG

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    Science Advisor
    Homework Helper

    The maximum of the sum of the squares isn't necessarily the maximum of the sum.
     
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