1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Minimizing Distance to the Origin
  1. Minimize Distance (Replies: 1)

Loading...