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Finding the absolute minimum of a function in R2

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Use technique of completing squares to Show that this function has an absolute minimum.
    f(x, y) = x^2 + y^2 − 2x + 4y + 1

    2. Relevant equations
    Not entirely sure how completing the squares will indicate an absolute minimum.Is there some additional reasoning required?

    3. The attempt at a solution
    Having completed the square i arrived at this equation.
    (x-1)^2 +(y+2)^2 = 4
    Again,all i can extrapolate from this is the center point of an obvious circle of radius 4.
     
  2. jcsd
  3. Nov 5, 2014 #2

    Ray Vickson

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    There is no way you could have arrived at the equation ##(x-1)^2 + (y-2)^2 = 4## if you had done correct work. Show your steps.
     
  4. Nov 5, 2014 #3

    Mark44

    Staff: Mentor

    But that doesn't answer the question.
    By completing the square, you can rewrite the formula for f(x, y). From that, it's not hard to show that this function has an absolute minimum.
     
    Last edited: Nov 5, 2014
  5. Nov 5, 2014 #4

    Dick

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    There is no equation here. You want the minimum of the function (x-1)^2+(y+2)^2-4. What might that be? How small can (x-1)^2 and (y+2)^2 be?
     
  6. Nov 5, 2014 #5
    Perhaps it was a typo on your part,because my solution was (x-1)^2 + (y+2)^2, I have done it several times.However if it is still incorrect i'm more than willing to attempt to complete the square again.
    The issue of using this new version of the function to find the absolute minimum, The first thing that comes to mind would be to substitute the values (0,0) into the function.Am I on the right path?
     
  7. Nov 5, 2014 #6

    Ray Vickson

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    The function f(x,y) is NOT equal to (x-1)^2 + (y+2)^2. What IS it equal to?
     
  8. Nov 5, 2014 #7
    Here are the steps i went through to arrive at my new version of f(x,y).
    Given :f(x, y) = x^2 + y^2 − 2x + 4y + 1
    By taking the half the coefficients of (-2x) and (4y) then squaring them to find the missing values for (__) within the brackets.
    (x^2 - 2x + __ )+ (y^2 + 4y + __) = -1

    which gave me

    (x^2 -2x +1) + (y^2 +4y +4) = -1

    Then adding both 1 and 4 to the LHS of the equation to get

    (x^2 -2x +1) + (y^2 +4y +4) = -1 + 1 + 4
    thus
    (x - 1)^2 + (y+2)^2 =4

    where did i go wrong?
     
  9. Nov 5, 2014 #8

    Mark44

    Staff: Mentor

    The step above is wrong. You are tacitly assuming that f(x, y) = 0 to get this equation. There is no reason to make this assumption. Work with the expression x2 + y2 - 2x + 4y + 1. When you add something to the x- or y-terms, subtract the same amount to get a new expression that is equal to the one you started with.

     
  10. Nov 5, 2014 #9
    Taking what you said into consideration;
    given f(x,y): x^2 -2x +1 +y^2 +4y

    The completed square should look like this -> f(x,y)= (x-1)^2 +(y+2)^2 - 4

    Assuming this holds,
    intuitively it seems the smallest possible value would occur at (1,-2). which results in -4 being the absolute minimum.?
     
  11. Nov 5, 2014 #10

    Mark44

    Staff: Mentor

    Yes.
     
  12. Nov 5, 2014 #11
    Thank you all.
     
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