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alffer1
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?
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\[d=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]
Oh my... *Pythagoras
ok for that I get sqrt(5)/3. Now how do I get the farthest point?
you didn't get the critical points yet it seems. actually I think they will be boundaries... let's see...
ok now I got myself totally confused...
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\[\\\begin{align}&\nabla f(x)&=\lambda\nabla g(x)\\(1)&8x&=2\lambda x\\(2)&2y-4&=2\lambda y\\(3)&2z-4&=2\lambda z\\(4)&4x^2+(y-2)^2+(z-1)&=4\end{align}\]
wait...I thought the Lambda was put on the side of the constraint derivatives...
yep, my bad, let me fix that :P
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
and I solved that and got the 4 values shown above...
so what do those 4 values mean right now?
they are a set of critical points, but I'm not sure they are the only ones...
well, just one critical point I mean
lambda means nothing as I mentioned before
But if there is only one critical point, how can there be a max distance AND a min distance?
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.
well if you plug x=0 into the restraint you get a circle...
ohhhhhhhh I think I see the problem again it is a matter of the ambiguity in the sign...
how did you solve for y ?
haha... plugged in lambda=1/4
um...I plugged in L = 1/4, got 2y = (y/2) -1
-1/4 would work for the x equation too, huh?
Where on earth did the negative come from?
since x is 0, the sign of Lambda doesn't matter...
maybe/ maybe not... but I think that when you divide out the x, you eliminate solutions.
Not to confuse the discussion, but I think there are two maximums.
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:
well we need to deal with L first
L=1/4 so 64=y^2+z^2=8^2
now we need a way to represent y in terms of z or vice versa
or not... but that's the plan
ok I'm working this out...
yeah that works\[\lambda={y\over y-1}={z\over z-1}\]
I think that might be y-2 instead of y-1
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!
I was just looking at that... thank goodness you saw it too.
wait, now are we supposed to solve for y and z?
yeah, this is crazy. I've filled pages with mostly garbage.
that's what i get, so plugging that into the restraint gives you...?
z = sqrt(20)/20 y = sqrt(20)/10
those are both positive negative by the way
yeah that's what I get, though I am disturbed that wolfram disagrees
oh I see, we should have just plugged this directly into the original restraint with x=0 4=(y-2)^2+(z-1)^2
oh oh ok that makes more sense
went a little overboard trying to simplify I guess
ok gotta go. Thanks so much!
with the +/- you have your max and min of course welcome!
Thanks for brushing out some cobwebs... more to go.
ikr? seems like history...