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alffer1

  • one year ago

Multivariable Help: Once again I'm stumbling on Lagrange Multipliers. I need to find the maximum and minimum distance from the origin of a point on the ellipsoid 4x^2 + (y-2)^2 + (z-1)^2 = 4. I've determined that the partials are as follows: df/dx = 2x df/dy = 2y df/dz = 2z dg/dx = 8x dg/dy = 2(y-2) dg/dz = 2(z-1) From this, I get the Lagrange equations to be: 2x = L(8x) 2y = L(2y-4) 2z = L(2z-2) From these equations, I get x = 0, L = 1/4, y = -2/3 and z = -1/3. Where do I go from here?

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  1. EulerGroupie
    • one year ago
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    I'm a little rusty, but it looks like you found the closest point to the origin that is on the ellipsoid. So follow Pythagorus\[d=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]

  2. EulerGroupie
    • one year ago
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    Oh my... *Pythagoras

  3. alffer1
    • one year ago
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    ok for that I get sqrt(5)/3. Now how do I get the farthest point?

  4. TuringTest
    • one year ago
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    you didn't get the critical points yet it seems. actually I think they will be boundaries... let's see...

  5. alffer1
    • one year ago
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    ok now I got myself totally confused...

  6. TuringTest
    • one year ago
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    yeah you did confuse things a tad... constraint:\[f(x)=4x^2+(y-2)^2+(z-1)^2=4\]to optimize we'll use distance squared\[g(x)=x^2+y^2+z^2\]we then apply Lagrange's trick to get\[\\\begin{align}&\nabla f(x)&=\lambda\nabla g(x)\\(1)&8x&=2\lambda x\\(2)&2y-4&=2\lambda y\\(3)&2z-4&=2\lambda z\\(4)&4x^2+(y-2)^2+(z-1)&=4\end{align}\]

  7. alffer1
    • one year ago
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    wait...I thought the Lambda was put on the side of the constraint derivatives...

  8. TuringTest
    • one year ago
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    yep, my bad, let me fix that :P

  9. TuringTest
    • one year ago
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    ok the heck with using the latex, takes too long you had it right: 2x = L(8x) 2y = L(2y-4) 2z = L(2z-2) 4x^2+(y-2)^2+(z-1)^2=4 it's a matter of solving this system: 4 equations, 4 unknowns

  10. alffer1
    • one year ago
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    and I solved that and got the 4 values shown above...

  11. alffer1
    • one year ago
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    so what do those 4 values mean right now?

  12. TuringTest
    • one year ago
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    they are a set of critical points, but I'm not sure they are the only ones...

  13. TuringTest
    • one year ago
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    well, just one critical point I mean

  14. TuringTest
    • one year ago
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    lambda means nothing as I mentioned before

  15. alffer1
    • one year ago
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    But if there is only one critical point, how can there be a max distance AND a min distance?

  16. EulerGroupie
    • one year ago
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    I have this tiny little voice in my head that keeps saying, "go through the constraint". I know it sounds weird, but I remember something about how there is always another one through the constraint.

  17. TuringTest
    • one year ago
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    well if you plug x=0 into the restraint you get a circle...

  18. TuringTest
    • one year ago
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    ohhhhhhhh I think I see the problem again it is a matter of the ambiguity in the sign...

  19. TuringTest
    • one year ago
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    how did you solve for y ?

  20. EulerGroupie
    • one year ago
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    haha... plugged in lambda=1/4

  21. alffer1
    • one year ago
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    um...I plugged in L = 1/4, got 2y = (y/2) -1

  22. EulerGroupie
    • one year ago
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    -1/4 would work for the x equation too, huh?

  23. alffer1
    • one year ago
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    Where on earth did the negative come from?

  24. alffer1
    • one year ago
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    oh I see what you mean

  25. alffer1
    • one year ago
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    since x is 0, the sign of Lambda doesn't matter...

  26. EulerGroupie
    • one year ago
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    maybe/ maybe not... but I think that when you divide out the x, you eliminate solutions.

  27. EulerGroupie
    • one year ago
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    Not to confuse the discussion, but I think there are two maximums.

  28. TuringTest
    • one year ago
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    see, you need to plug in that x value into the restraint, using that value of lambda 2x = L(8x) -> L=1/4, x=0 2y = L(2y-4) -> y-2=y/L 2z = L(2z-2) -> z-1=z/L plug these into 4x^2+(y-2)^2+(z-1)^2=4 now look how we can rewrite the restraint:

  29. TuringTest
    • one year ago
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    4=(y/L)^2+(z/L)^2

  30. alffer1
    • one year ago
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    circle of radius 2 ?

  31. TuringTest
    • one year ago
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    well we need to deal with L first

  32. TuringTest
    • one year ago
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    L=1/4 so 64=y^2+z^2=8^2

  33. TuringTest
    • one year ago
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    now we need a way to represent y in terms of z or vice versa

  34. TuringTest
    • one year ago
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    or not... but that's the plan

  35. alffer1
    • one year ago
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    ok I'm working this out...

  36. TuringTest
    • one year ago
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    me too :P

  37. TuringTest
    • one year ago
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    yeah that works\[\lambda={y\over y-1}={z\over z-1}\]

  38. EulerGroupie
    • one year ago
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    I think that might be y-2 instead of y-1

  39. TuringTest
    • one year ago
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    oh yes, sorry!

  40. TuringTest
    • one year ago
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    also it should be y^2+z^2=1/4 after multiplying by L^2 this is why we always need extra eyes on these things!

  41. EulerGroupie
    • one year ago
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    I was just looking at that... thank goodness you saw it too.

  42. alffer1
    • one year ago
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    wait, now are we supposed to solve for y and z?

  43. EulerGroupie
    • one year ago
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    yeah, this is crazy. I've filled pages with mostly garbage.

  44. alffer1
    • one year ago
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    oh y = 2z haha

  45. TuringTest
    • one year ago
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    that's what i get, so plugging that into the restraint gives you...?

  46. alffer1
    • one year ago
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    z = sqrt(20)/20 y = sqrt(20)/10

  47. alffer1
    • one year ago
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    those are both positive negative by the way

  48. alffer1
    • one year ago
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    is that right?

  49. TuringTest
    • one year ago
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    yeah that's what I get, though I am disturbed that wolfram disagrees

  50. alffer1
    • one year ago
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    oh boy...

  51. TuringTest
    • one year ago
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    oh I see, we should have just plugged this directly into the original restraint with x=0 4=(y-2)^2+(z-1)^2

  52. alffer1
    • one year ago
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    oh oh ok that makes more sense

  53. TuringTest
    • one year ago
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    went a little overboard trying to simplify I guess

  54. alffer1
    • one year ago
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    ok gotta go. Thanks so much!

  55. TuringTest
    • one year ago
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    with the +/- you have your max and min of course welcome!

  56. EulerGroupie
    • one year ago
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    Thanks for brushing out some cobwebs... more to go.

  57. TuringTest
    • one year ago
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    ikr? seems like history...

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