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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 + (y2)^2 + (z1)^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(y2)
dg/dz = 2(z1)
From this, I get the Lagrange equations to be:
2x = L(8x)
2y = L(2y4)
2z = L(2z2)
From these equations, I get x = 0, L = 1/4, y = 2/3 and z = 1/3. Where do I go from here?
 4 months ago
 4 months 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 + (y2)^2 + (z1)^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(y2) dg/dz = 2(z1) From this, I get the Lagrange equations to be: 2x = L(8x) 2y = L(2y4) 2z = L(2z2) From these equations, I get x = 0, L = 1/4, y = 2/3 and z = 1/3. Where do I go from here?
 4 months ago
 4 months ago

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EulerGroupieBest ResponseYou've already chosen the best response.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\[d=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
Oh my... *Pythagoras
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
ok for that I get sqrt(5)/3. Now how do I get the farthest point?
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
you didn't get the critical points yet it seems. actually I think they will be boundaries... let's see...
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
ok now I got myself totally confused...
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
yeah you did confuse things a tad... constraint:\[f(x)=4x^2+(y2)^2+(z1)^2=4\]to optimize we'll use distance squared\[g(x)=x^2+y^2+z^2\]we then apply Lagrange's trick to get\[\\\begin{align}&\nabla f(x)&=\lambda\nabla g(x)\\(1)&8x&=2\lambda x\\(2)&2y4&=2\lambda y\\(3)&2z4&=2\lambda z\\(4)&4x^2+(y2)^2+(z1)&=4\end{align}\]
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
wait...I thought the Lambda was put on the side of the constraint derivatives...
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
yep, my bad, let me fix that :P
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
ok the heck with using the latex, takes too long you had it right: 2x = L(8x) 2y = L(2y4) 2z = L(2z2) 4x^2+(y2)^2+(z1)^2=4 it's a matter of solving this system: 4 equations, 4 unknowns
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
and I solved that and got the 4 values shown above...
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
so what do those 4 values mean right now?
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
they are a set of critical points, but I'm not sure they are the only ones...
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
well, just one critical point I mean
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
lambda means nothing as I mentioned before
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
But if there is only one critical point, how can there be a max distance AND a min distance?
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
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.
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
well if you plug x=0 into the restraint you get a circle...
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
ohhhhhhhh I think I see the problem again it is a matter of the ambiguity in the sign...
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
how did you solve for y ?
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
haha... plugged in lambda=1/4
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
um...I plugged in L = 1/4, got 2y = (y/2) 1
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
1/4 would work for the x equation too, huh?
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
Where on earth did the negative come from?
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
oh I see what you mean
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
since x is 0, the sign of Lambda doesn't matter...
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
maybe/ maybe not... but I think that when you divide out the x, you eliminate solutions.
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
Not to confuse the discussion, but I think there are two maximums.
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.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(2y4) > y2=y/L 2z = L(2z2) > z1=z/L plug these into 4x^2+(y2)^2+(z1)^2=4 now look how we can rewrite the restraint:
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
well we need to deal with L first
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
L=1/4 so 64=y^2+z^2=8^2
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
now we need a way to represent y in terms of z or vice versa
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
or not... but that's the plan
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
ok I'm working this out...
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
yeah that works\[\lambda={y\over y1}={z\over z1}\]
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
I think that might be y2 instead of y1
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.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!
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
I was just looking at that... thank goodness you saw it too.
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
wait, now are we supposed to solve for y and z?
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
yeah, this is crazy. I've filled pages with mostly garbage.
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
that's what i get, so plugging that into the restraint gives you...?
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
z = sqrt(20)/20 y = sqrt(20)/10
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
those are both positive negative by the way
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
yeah that's what I get, though I am disturbed that wolfram disagrees
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
oh I see, we should have just plugged this directly into the original restraint with x=0 4=(y2)^2+(z1)^2
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
oh oh ok that makes more sense
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
went a little overboard trying to simplify I guess
 4 months ago

alffer1Best ResponseYou've already chosen the best response.1
ok gotta go. Thanks so much!
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
with the +/ you have your max and min of course welcome!
 4 months ago

EulerGroupieBest ResponseYou've already chosen the best response.1
Thanks for brushing out some cobwebs... more to go.
 4 months ago

TuringTestBest ResponseYou've already chosen the best response.2
ikr? seems like history...
 4 months ago
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