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anonymous
 3 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 + (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?
anonymous
 3 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 + (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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm a little rusty, but it looks like you found the closest point to the origin that is on the ellipsoid. So follow Pythagorus\[d=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok for that I get sqrt(5)/3. Now how do I get the farthest point?

TuringTest
 3 years 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok now I got myself totally confused...

TuringTest
 3 years 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\[\\\begin{align}&\nabla f(x)&=\lambda\nabla g(x)\\(1)&8x&=2\lambda x\\(2)&2y4&=2\lambda y\\(3)&2z4&=2\lambda z\\(4)&4x^2+(y2)^2+(z1)&=4\end{align}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait...I thought the Lambda was put on the side of the constraint derivatives...

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

TuringTest
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and I solved that and got the 4 values shown above...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what do those 4 values mean right now?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But if there is only one critical point, how can there be a max distance AND a min distance?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.2well if you plug x=0 into the restraint you get a circle...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha... plugged in lambda=1/4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0um...I plugged in L = 1/4, got 2y = (y/2) 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01/4 would work for the x equation too, huh?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Where on earth did the negative come from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh I see what you mean

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since x is 0, the sign of Lambda doesn't matter...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe/ maybe not... but I think that when you divide out the x, you eliminate solutions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not to confuse the discussion, but I think there are two maximums.

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok I'm working this out...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think that might be y2 instead of y1

TuringTest
 3 years 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was just looking at that... thank goodness you saw it too.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, now are we supposed to solve for y and z?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, this is crazy. I've filled pages with mostly garbage.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0z = sqrt(20)/20 y = sqrt(20)/10

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0those are both positive negative by the way

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

TuringTest
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh oh ok that makes more sense

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok gotta go. Thanks so much!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for brushing out some cobwebs... more to go.

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