A community for students.
Here's the question you clicked on:
 0 viewing
alffer1
 2 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?
alffer1
 2 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?

This Question is Closed

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

alffer1
 2 years 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
 2 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...

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

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

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

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

TuringTest
 2 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

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

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

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

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

alffer1
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.2well if you plug x=0 into the restraint you get a circle...

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

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

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

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

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

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

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

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

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

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

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

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

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

TuringTest
 2 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!

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

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

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

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

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

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

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

TuringTest
 2 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

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

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

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

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2ikr? seems like history...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.