## anonymous 2 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?

1. anonymous

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

Oh my... *Pythagoras

3. anonymous

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

4. TuringTest

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

5. anonymous

ok now I got myself totally confused...

6. TuringTest

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

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

8. TuringTest

yep, my bad, let me fix that :P

9. TuringTest

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

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

11. anonymous

so what do those 4 values mean right now?

12. TuringTest

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

13. TuringTest

well, just one critical point I mean

14. TuringTest

lambda means nothing as I mentioned before

15. anonymous

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

16. anonymous

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

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

18. TuringTest

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

19. TuringTest

how did you solve for y ?

20. anonymous

haha... plugged in lambda=1/4

21. anonymous

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

22. anonymous

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

23. anonymous

Where on earth did the negative come from?

24. anonymous

oh I see what you mean

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. TuringTest

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

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

30. anonymous

circle of radius 2 ?

31. TuringTest

well we need to deal with L first

32. TuringTest

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

33. TuringTest

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

34. TuringTest

or not... but that's the plan

35. anonymous

ok I'm working this out...

36. TuringTest

me too :P

37. TuringTest

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

38. anonymous

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

39. TuringTest

oh yes, sorry!

40. TuringTest

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

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

42. anonymous

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

43. anonymous

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

44. anonymous

oh y = 2z haha

45. TuringTest

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

46. anonymous

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

47. anonymous

those are both positive negative by the way

48. anonymous

is that right?

49. TuringTest

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

50. anonymous

oh boy...

51. TuringTest

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

oh oh ok that makes more sense

53. TuringTest

went a little overboard trying to simplify I guess

54. anonymous

ok gotta go. Thanks so much!

55. TuringTest

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

56. anonymous

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

57. TuringTest

ikr? seems like history...