Question

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?

Answer 1

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\[d=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]

Answer 2

Oh my... *Pythagoras

Answer 3

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

Answer 4

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

Answer 5

ok now I got myself totally confused...

Answer 6

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\[\\\begin{align}&\nabla f(x)&=\lambda\nabla g(x)\\(1)&8x&=2\lambda x\\(2)&2y-4&=2\lambda y\\(3)&2z-4&=2\lambda z\\(4)&4x^2+(y-2)^2+(z-1)&=4\end{align}\]

Answer 7

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

Answer 8

yep, my bad, let me fix that :P

Answer 9

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

Answer 10

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

Answer 11

so what do those 4 values mean right now?

Answer 12

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

Answer 13

well, just one critical point I mean

Answer 14

lambda means nothing as I mentioned before

Answer 15

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

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

how did you solve for y ?

Answer 20

haha... plugged in lambda=1/4

Answer 21

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

Answer 22

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

Answer 23

Where on earth did the negative come from?

Answer 24

oh I see what you mean

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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:

Answer 29

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

Answer 30

circle of radius 2 ?

Answer 31

well we need to deal with L first

Answer 32

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

Answer 33

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

Answer 34

or not... but that's the plan

Answer 35

ok I'm working this out...

Answer 36

me too :P

Answer 37

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

Answer 38

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

Answer 39

oh yes, sorry!

Answer 40

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!

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

oh y = 2z haha

Answer 45

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

Answer 46

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

Answer 47

those are both positive negative by the way

Answer 48

is that right?

Answer 49

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

Answer 50

oh boy...

Answer 51

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

Answer 52

oh oh ok that makes more sense

Answer 53

went a little overboard trying to simplify I guess

Answer 54

ok gotta go. Thanks so much!

Answer 55

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

Answer 56

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

Answer 57

ikr? seems like history...