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Find the points closest to the origin

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the points [itex]x^2 + xy + y^2 = 2[/itex] that are closest to the origin.


    2. Relevant equations
    Distance Formula


    3. The attempt at a solution
    I have to first solve this without using Lagrange Multipliers.

    This is essentially an ellipse. So I first completed the square:

    [itex]3/4\,{x}^{2}+ \left( y+1/2\,x \right) ^{2}=2[/itex]

    I was thinking first I should separate [itex]x[/itex] and [itex]y[/itex], and use that in the distance formula, but I can't seem to isolate [itex]y[/itex], is my approach wrong?
     
    Last edited: Apr 25, 2012
  2. jcsd
  3. Apr 25, 2012 #2

    Mark44

    Staff: Mentor

    Shouldn't this be an equation?
    Where did the 2 come from? Are you starting with ##x^2 + xy + y^2 = 2##?
     
  4. Apr 25, 2012 #3
    Yes, sorry. That was silly of me, I edited the original post to correct that error.
     
  5. Apr 25, 2012 #4

    Mark44

    Staff: Mentor

    The equation y2 + xy + x2 - 2 = 0 is quadratic in y, so you can use the Quadratic Formula to solve for y. (You'll get two values with the ±, meaning there are two functions of x.)

    Your problem is to find the minimum value of D(x) = √(x2 + y2). Here you will need to substitute for y from the previous work, making D really a function of x alone.

    Alternatively, you could find the minimum value of D2(x) = x2 + y2, again making the substitutions for y. Since there are two functions, you'll need to do the work for both of them. I don't know if there is any symmetry you can exploit to save work. The graph is an ellipse, but one that has been rotated.
     
  6. Apr 25, 2012 #5
    Thanks for that, I will look into it.
     
  7. Apr 26, 2012 #6

    Ray Vickson

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    As an alternative method, you could solve the problem as a constrained optimization, using a Lagrange multiplier approach. You want to minimize f(x,y) = x^2 + y^2 (the square of the distance) subject to g(x,y) = 0, where g(x,y) = x^2 + y^2 + x*y-2.

    RGV
     
  8. Apr 26, 2012 #7

    Mark44

    Staff: Mentor

    In post #1 it says
     
  9. Apr 26, 2012 #8

    D H

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    Another option is to use polar coordinates. The equation of the ellipse becomes [itex]r^2(1+\sin\theta\cos\theta) = 2[/itex], where [itex]r[/itex] is the distance from the origin.

    And that's a big enough hint at this point.
     
  10. Apr 26, 2012 #9

    Ray Vickson

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    I missed that remark about no Lagrange multipliers, although I cannot see why anyone would say "you cannot use such-and-such a technique". What's the point of that? However, the OP must live with it.

    RGV
     
  11. Apr 26, 2012 #10

    D H

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    It's probably because the next part of the problem (not shown here) was to solve the same problem, but this time using Lagrange multipliers.
     
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