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alffer1
 one year 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?
alffer1
 one year 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?

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EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1I'm a little rusty, but it looks like you found the closest point to the origin that is on the ellipsoid. So follow Pythagorus\[d=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1Oh my... *Pythagoras

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1ok for that I get sqrt(5)/3. Now how do I get the farthest point?

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

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1ok now I got myself totally confused...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2yeah 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\[\\\begin{align}&\nabla f(x)&=\lambda\nabla g(x)\\(1)&8x&=2\lambda x\\(2)&2y4&=2\lambda y\\(3)&2z4&=2\lambda z\\(4)&4x^2+(y2)^2+(z1)&=4\end{align}\]

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1wait...I thought the Lambda was put on the side of the constraint derivatives...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2yep, my bad, let me fix that :P

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2ok 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

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1and I solved that and got the 4 values shown above...

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1so what do those 4 values mean right now?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2they are a set of critical points, but I'm not sure they are the only ones...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2well, just one critical point I mean

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2lambda means nothing as I mentioned before

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1But if there is only one critical point, how can there be a max distance AND a min distance?

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2well if you plug x=0 into the restraint you get a circle...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2ohhhhhhhh I think I see the problem again it is a matter of the ambiguity in the sign...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2how did you solve for y ?

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1haha... plugged in lambda=1/4

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1um...I plugged in L = 1/4, got 2y = (y/2) 1

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.11/4 would work for the x equation too, huh?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1Where on earth did the negative come from?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1oh I see what you mean

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1since x is 0, the sign of Lambda doesn't matter...

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1maybe/ maybe not... but I think that when you divide out the x, you eliminate solutions.

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1Not to confuse the discussion, but I think there are two maximums.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2see, 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:

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2well we need to deal with L first

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2L=1/4 so 64=y^2+z^2=8^2

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2now we need a way to represent y in terms of z or vice versa

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2or not... but that's the plan

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1ok I'm working this out...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2yeah that works\[\lambda={y\over y1}={z\over z1}\]

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1I think that might be y2 instead of y1

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2also 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!

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1I was just looking at that... thank goodness you saw it too.

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1wait, now are we supposed to solve for y and z?

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1yeah, this is crazy. I've filled pages with mostly garbage.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2that's what i get, so plugging that into the restraint gives you...?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1z = sqrt(20)/20 y = sqrt(20)/10

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1those are both positive negative by the way

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2yeah that's what I get, though I am disturbed that wolfram disagrees

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

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1oh oh ok that makes more sense

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2went a little overboard trying to simplify I guess

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1ok gotta go. Thanks so much!

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2with the +/ you have your max and min of course welcome!

EulerGroupie
 one year ago
Best ResponseYou've already chosen the best response.1Thanks for brushing out some cobwebs... more to go.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2ikr? seems like history...
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