A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 3 years 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?

  • This Question is Closed
  1. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh my... *Pythagoras

  3. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok for that I get sqrt(5)/3. Now how do I get the farthest point?

  4. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you didn't get the critical points yet it seems. actually I think they will be boundaries... let's see...

  5. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok now I got myself totally confused...

  6. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait...I thought the Lambda was put on the side of the constraint derivatives...

  8. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yep, my bad, let me fix that :P

  9. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and I solved that and got the 4 values shown above...

  11. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so what do those 4 values mean right now?

  12. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    they are a set of critical points, but I'm not sure they are the only ones...

  13. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well, just one critical point I mean

  14. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    lambda means nothing as I mentioned before

  15. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But if there is only one critical point, how can there be a max distance AND a min distance?

  16. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well if you plug x=0 into the restraint you get a circle...

  18. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ohhhhhhhh I think I see the problem again it is a matter of the ambiguity in the sign...

  19. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    how did you solve for y ?

  20. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha... plugged in lambda=1/4

  21. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    um...I plugged in L = 1/4, got 2y = (y/2) -1

  22. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -1/4 would work for the x equation too, huh?

  23. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where on earth did the negative come from?

  24. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh I see what you mean

  25. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since x is 0, the sign of Lambda doesn't matter...

  26. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe/ maybe not... but I think that when you divide out the x, you eliminate solutions.

  27. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Not to confuse the discussion, but I think there are two maximums.

  28. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    4=(y/L)^2+(z/L)^2

  30. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    circle of radius 2 ?

  31. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well we need to deal with L first

  32. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    L=1/4 so 64=y^2+z^2=8^2

  33. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now we need a way to represent y in terms of z or vice versa

  34. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    or not... but that's the plan

  35. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok I'm working this out...

  36. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    me too :P

  37. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah that works\[\lambda={y\over y-1}={z\over z-1}\]

  38. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think that might be y-2 instead of y-1

  39. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh yes, sorry!

  40. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was just looking at that... thank goodness you saw it too.

  42. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, now are we supposed to solve for y and z?

  43. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah, this is crazy. I've filled pages with mostly garbage.

  44. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh y = 2z haha

  45. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that's what i get, so plugging that into the restraint gives you...?

  46. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    z = sqrt(20)/20 y = sqrt(20)/10

  47. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    those are both positive negative by the way

  48. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is that right?

  49. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah that's what I get, though I am disturbed that wolfram disagrees

  50. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh boy...

  51. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh oh ok that makes more sense

  53. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    went a little overboard trying to simplify I guess

  54. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok gotta go. Thanks so much!

  55. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    with the +/- you have your max and min of course welcome!

  56. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks for brushing out some cobwebs... more to go.

  57. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ikr? seems like history...

  58. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.