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Math Help - Can anyone help me to solve this problem. Thank.

  1. #1
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    Can anyone help me to solve this problem. Thank.

    please help me on the problem of finding minimum value. Thank.

    http://www.math.um.edu.my/Ismweb/Ome.../Problem+1.pdf
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SuperTu
    please help me on the problem of finding minimum value. Thank.

    http://www.math.um.edu.my/Ismweb/Ome.../Problem+1.pdf
    y = \sqrt{(x+a)^2+b} + \sqrt{(x+c)^2+d}

    I presume you mean the minimum value of y?

    Set \frac{dy}{dx} = \frac{2(x+a)}{\sqrt{(x+a)^2+b}} + \frac{2(x+c)}{\sqrt{(x+c)^2+d}} = 0

    Find a common denominator:
    \frac{2(x+a)\sqrt{(x+c)^2+d} + 2(x+c)\sqrt{(x+a)^2+b}}{\sqrt{(x+a)^2+b} \sqrt{(x+c)^2+d}}=0

    Thus:
    2(x+a)\sqrt{(x+c)^2+d} + 2(x+c)\sqrt{(x+a)^2+b} = 0

    (x+a)\sqrt{(x+c)^2+d} = -(x+c)\sqrt{(x+a)^2+b}

    Square both sides:
    (x^2 + 2ax + a^2) (x^2 + 2cx + c^2 + d) = (x^2 + 2cx + c^2) (x^2 + 2ax + a^2 + b)

    If you expand this on both sides of the equation you will find that you are left with a quadratic (there are a number of nice cancellations) that you can solve for x. Then put this value of x into the y equation to find a value of y. (Make sure that the final value you claim is the answer is, in fact, a minimum not a maximum.)

    A quicker way to do this is logical. You are adding two positive terms to find a minimum, so find the minimum values of \sqrt{(x+a)^2+b} and \sqrt{(x+c)^2+d}. Then add these to get the minimum value for y.

    -Dan
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  3. #3
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    There is a much more simplified version.
    <br />
y = \sqrt{(x+a)^2+b} + \sqrt{(x+c)^2+d}<br />
    If and only if,
    y=(x+a)^2+b+(x+c)^2+d
    (This is the same idea you use in the Method of Least squares).

    Anyway, you can do it like topsquark said with the derivative but you can even do it without it.
    Open,
    y=x^2+2ax+a^2+b+x^2+2cx+c^2+d
    Organize,
    y=2x^2+(2a+2c)x+(a^2+c^2+b+d)
    This is parabola with a minimum point which occurs at,
    x=-\frac{2a+2b}{4}=-1/2(a+b)
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  4. #4
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    you all didn't see the question clearly

    you all didn't see the question clearly.. there should be b square and d square in that equation but not b anf d. thank
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SuperTu
    you all didn't see the question clearly.. there should be b square and d square in that equation but not b anf d. thank
    My apologies for reading it wrong, but the method of solution is identical.

    -Dan
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  6. #6
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    may i know what theorem or statment should be used to explain that the stationary is minimum point instead of maximum or inflection point. thank
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SuperTu
    may i know what theorem or statment should be used to explain that the stationary is minimum point instead of maximum or inflection point. thank
    In general there isn't one. Personally I would use the characteristics of the graph of the function to decide if we have a minimum point there, but in general you need to use the second derivative test:

    If \frac{d^2y}{dx^2}<0 then the stationary point is a relative maximum.

    If \frac{d^2y}{dx^2}>0 then the stationary point is a relative minimum.

    If \frac{d^2y}{dx^2}=0 then the stationary point is a point of inflection.

    I mentioned that we need to check this near the bottom of my first post.

    -Dan
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  8. #8
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    Let me add something.
    \frac{d^2y}{dx^2}=0--->Inflection point.
    But you know nothing about whether it is maximum of minimum! You need to return back to the first derivative test. A common error is that people think whenever the second derivative is zero it means it is neither a minimum nor a maximum (I think the name of the term is 'saddle point')
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  9. #9
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    Hello, Super!

    Given that a,b,c,d are real numbers,

    find the minimum value of: . y\;=\;\sqrt{(x+a) + b^2} + \sqrt{(x+c)^2+d^2}

    Consider the following problem . . .

    Given two points Q(-a,b) and R(-c,d), find the point P(x,0) on the x-axis
    . . such that the sum of the distances \overline{PQ} + \overline{PR} is a minimum.
    Code:
                    R       |
                    *(-c,d) |
          Q        /        |
    (-a,b)*       /         |
           \     /          |
            \   /           |
             \ /            |
          - - * - - - - - - + - - -
              P
            (x,0)

    Since PQ\:=\:\sqrt{(x -[-a])^2 + (0 - b)^2} \:=\:\sqrt{(x + a)^2 + b^2}

    . and PR\:=\:\sqrt{(x - [-c])^2 + (0 - d)^2} \;= \;\sqrt{(x + c)^2 + d^2}

    then: . S\:=\:\sqrt{(x+a)^2 + b^2} + \sqrt{(x+c)^2 + d)^2}

    . . and we have a geometric interpretation of your problem.

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  10. #10
    Grand Panjandrum
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    Will you cut us in for a share of the mystery gift if you win it?

    I don't think you should be asking for our help, nor should we be giving it.

    This link tells us about the context of this problem

    RonL
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  11. #11
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    Quote Originally Posted by Soroban
    ...

    Given two points Q(-a,b) and R(-c,d), find the point P(x,0) on the x-axis
    . . such that the sum of the distances \overline{PQ} + \overline{PR} is a minimum.
    Code:
                    R       |
                    *(-c,d) |
          Q        /        |
    (-a,b)*       /         |
           \     /          |
            \   /           |
             \ /            |
          - - * - - - - - - + - - -
              P
            (x,0)
    ...
    Hello, Soroban,

    your explanation of the situation and especially your sketch tickled my memory: This problem was solved by Archimedes already - if I remember correctly. (Reflect R at the x-Axis. The straight line between Q and R' isthe shortest distance between Q and R. P is the zero of this line)

    Greetings

    EB
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