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Homework Help: Minimizing Distance to the Origin

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the point(s) on the surface z2 - 10xy = 10 nearest to the origin.

    3. The attempt at a solution
    I will minimize distance by minimizing distance2. So,

    D = x2 + y2 + z2, where z2 - 10xy = 10. By Lagrange multipliers I have,

    2x = -10yλ
    2y = -10xλ
    2z = 2zλ.

    By the last equation, clearly either z = 0 or λ = 1 or λ = 0.

    I will test λ = 1 first with the condition y ≠ 0 and x ≠ 0. By dividing the first equation by the second equation I have,

    x/y = y/x, which implies x = y. But clearly 2x ≠ -10x when x ≠ 0. So I reject λ = 1 with the condition y ≠ 0 and x ≠ 0. I will now test λ = 1 first with the condition y = 0. I have,

    2(0) = 10 * x * 1, so clearly x = 0. (Similarly, I could have tested with the condition x = 0 and found y = 0.) By the constraint equation I have z2 - 10(0)(0) = 10, so clearly z = ±√10.

    Suppose z = 0. Then x ≠ 0 and y ≠ 0 because, in our constraint equation, -10xy = 10. So clearly xy = -1. I will now divide 2x = -10yλ by 2y = -10xλ and again find x = y. This implies,

    x2 = -1 or y2 = -1.

    x = i or y = i. I reject these solutions as crazy.

    Suppose λ = 0. Then it must follow that y = 0 and x = 0. Again, plugging into our constraint equation gives z = ±√10. Therefore, Lagrange multipliers give the points (0,0,±√10), both of which give distance √10.

    However, by inspection, one can see that the points (1,-1,0) and (-1,1,0) also satisfy the constraint and give distance √2.

    Ideas where I went wrong?
     
  2. jcsd
  3. Nov 14, 2011 #2

    I like Serena

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    First, λ = 0 is not a solution from your 3rd equation. (Check it.)

    Second, your conclusion x=y is not quite right.
    What you have is x^2=y^2, which has 2 solutions: x=y or x=-y.
     
  4. Nov 14, 2011 #3
    Okay. You're right about λ = 0. My bad.

    Let's see if I can actually derive the correct (I presume, unless there is another point closer to the origin that satisfies the restraint) solutions.

    So, from x2 = y2 I determine that x = ± y.

    02 -10(±y)(y) = 10.
    (-/+)y2 = 1
    (-/+)y = ± 1
    y = (-/+) 1

    x = -y, so when y = -1, x = 1. When y = 1, x = -1. Therefore, I also have solutions (±1,(-/+)1,0), both of which give distance √2. Look about right?

    Why, however, do (±1,(-/+)1,0) fail to satisfy 2x = -10y*λ and 2y = -10x*λ, when λ = 1?
     
  5. Nov 14, 2011 #4

    I like Serena

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    When λ=1, you do not have x,y=±1.

    What you do have is x=±y.
    And from 2x = -10yλ, it follows (with λ=1) that x=y=0.


    Btw, your solutions (±1,(-/+)1,0), both of which give distance √2, look right.
     
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