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Homework Help: Shortest Distance

  1. Mar 14, 2009 #1


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    1. The problem statement, all variables and given/known data
    Find the shortest distance from the origin to the curve [tex]x^2+2xy+y^2=150[/tex].

    2. Relevant equations
    [tex]\frac{\partial f}{\partial x}[/tex], [tex]\frac{\partial f}{\partial y}[/tex]

    3. The problem I'm occuring
    I'm not sure how to start is thus can't attempt it. I would have used the Lagrange theory but that would give the max point and I also don't have the constraint so cannot use it. For a start, can someone suggest how I would start. Any methods? I have to use partial differentiation.
  2. jcsd
  3. Mar 14, 2009 #2
    You can use Lagrange's multiplier method to find maxima as well as minima. Apply it to the distance squared function f(x,y)=x^2+y^2 with the constraint x^2+2xy+y^2=150.
  4. Mar 15, 2009 #3


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    I have reached a point of confusion...

    [tex]f(x,y) = x^2+y^2[/tex]
    [tex]g(x,y) = x^2+2xy+y^2=150[/tex]

    [tex]\mathrm{Equation 1: }2x - \lambda (2x + 2y) = 0 \implies \lambda = \frac{2x}{2x+2y}[/tex]
    [tex]\mathrm{Equation 2: }2y - \lambda (2x + 2y) = 0 \rightarrow 2y - \left( \frac{2x}{2x+2y}\right) (2x + 2y ) = 0 \rightarrow 2x+2y = 0 \implies y=x[/tex]

    Into constraint equation to solve for value of [tex]x[/tex] and [tex]y[/tex].
    [tex]\therefore y^2 + 2y^2 y^2 = 150 \implies y = \pm \sqrt{\frac{150}{4}} = x[/tex]

    Which gives many solutions. How can I saw which is the shortest distance?
  5. Mar 15, 2009 #4
    Well, you got two solutions: [tex](\sqrt{150}/2,\sqrt{150}/2)[/tex] and [tex](-\sqrt{150}/2,-\sqrt{150}/2)[/tex]. Both have the same distance from the origin, compute it and you have the answer.
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