# Finding Specific Extrema when grad(F)=constant & Lagrange Gives y=-z/2

Homework Statement:
Consider a particle that moves in x^2+2y^2+2z^2<=4 and is subject to the force field f(x,y,z)= (1,-2z,-2y). If f=grad(F), find the values in this region that maximize f.
Relevant Equations:
Lagrange Mulitpliers Technique
I found that f= x -2yz. To maximize f, I can first inspect the solutions to grad(F)=0. z=y=0 pops out, but I'm not sure what to do with the x-component equaling 1. Do we just include (x,0,0) as a solution? I think the problem wants specifics though, based on what I've seen previously from problem sets by this instructor.

Using Lagrange Multipliers, we get:
1=2xλ
-2z=4yλ
-2y=4zλ

Working through this, I find that z(1-4λ)=0. If z=0, then y=0. If λ=1/4, then y=-z/2 and x=2. So it appears we have (x,0,0) (as before) and (2, -z/2, z).

I'm not sure how to work through these to get the specifics here. Am I missing something, or is this a departure from this instructor's typical problems? Also, I don't think I need to parameterize and look for points inside the boundary here because of the Laplacian being equal to zero (I read that somewhere, but could be mistaken).

BvU
Homework Helper
... f=grad(F), find the values in this region that maximize f.

Don't you think that is a little contradictory ? Why should f=0 be a maximum ?

Or is your notation so sloppy it is beyond deciphering ? How is a maximum of a vector defined ?

Don't you think that is a little contradictory ? Why should f=0 be a maximum ?

Or is your notation so sloppy it is beyond deciphering ? How is a maximum of a vector defined ?
Oops, grad(f)=0. All the work is correct consistent, I just must have held the shift button after hitting the parenthesis. In any case, the the derivatives are set equal to zero, which is I set (1,-2z,-2y)=0, then also did the Lagrange stuff.

BvU
Homework Helper
I set (1,-2z,-2y)=0
That is ##\vec f = \vec 0## ? Searching for a "maximum" of the vector f ?

Please render the complete and litteral problem statement, exactly as given to you.

BvU
Homework Helper
I just must have held the shift button after hitting the parenthesis
Several times, it seems to me. Aren't you looking for a maximum of F instead of f ?

Several times, it seems to me. Aren't you looking for a maximum of F instead of f ?
Attached is the question, and I'm working on part c. I'm maximizing f, and thus looking at F, because it's the gradient. My work was showing setting (1,-2z,-2y)=0, but I don't know what 1=0 implies for the x-component. Also, with Lagrange I don't get anything that creates definitive values, but I believe I need them.

Also, not sure why I'm being criticized repeatedly for a typo-- it happens #### Attachments

• Screen Shot 2020-08-12 at 11.11.41 AM.png
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BvU
Homework Helper
I'm not trying to burn you with criticism, I am trying to help.

You wrote
is subject to the force field f(x,y,z)= (1,-2z,-2y). If f=grad(F)
but meant to write is subject to the force field ##\ \vec F(x,y,z)= (1,-2z,-2y).\ ## If ##\ \vec F=\vec\nabla (f) ##...

You found, correctly, ##f = x-2yz##.

What did you do to show that ##\vec F ## is conservative ?

In part (b) you found the ##W## that @etotheipi mentions, and finding extrema for that amounts to finding extrema for ##f##.

Lagrange multipliers work for constraints that can be written as equalities. Make an educated guess and search on ##x^2+2y^2+2z^2\ {\bf =} \ 4\;##. Check later.

(afraid to add insult to injury ):
... To maximize f, I can first inspect the solutions to grad(F)=0. z=y=0 pops out, but I'm not sure what to do with the x-component equaling 1. Do we just include (x,0,0) as a solution?
I don't follow at all. ##\vec \nabla f = (1,0,0)## not zero anywhere.

Anyway, you have the constraint (plus the guess) and want to tackle
Using Lagrange Multipliers, we get:
1=2xλ
-2z=4yλ
-2y=4zλ
But that is four unknowns and three equations. You need a fourth equation: the constraint !

Working through this, I find that z(1-4λ)=0. If z=0, then y=0. If λ=1/4, then y=-z/2 and x=2. So it appears we have (x,0,0) (as before) and (2, -z/2, z).
I get ##\ z(1-4\lambda^2) = 0\ ## (or did I just bump into another typo )

Keep going !

## \ ##

• I'm not trying to burn you with criticism, I am trying to help.

You wrote
but meant to write is subject to the force field ##\ \vec F(x,y,z)= (1,-2z,-2y).\ ## If ##\ \vec F=\vec\nabla (f) ##...

You found, correctly, ##f = x-2yz##.

What did you do to show that ##\vec F ## is conservative ?

In part (b) you found the ##W## that @etotheipi mentions, and finding extrema for that amounts to finding extrema for ##f##.

Lagrange multipliers work for constraints that can be written as equalities. Make an educated guess and search on ##x^2+2y^2+2z^2\ {\bf =} \ 4\;##. Check later.

(afraid to add insult to injury ):
I don't follow at all. ##\vec \nabla f = (1,0,0)## not zero anywhere.

Anyway, you have the constraint (plus the guess) and want to tackle

But that is four unknowns and three equations. You need a fourth equation: the constraint !

I get ##\ z(1-4\lambda^2) = 0\ ## (or did I just bump into another typo )

Keep going !

## \ ##
Haha, yeah, I definitely made a mistake there. λ=1/2 or -1/2. Thanks for the hint on adding the constraint, that makes sense and helps pull it together conceptually.

OK so, I have a few paths: λ=1/2 implies y=-z and x=1, so I get (1, -z, z), whereas λ=-1/2 implies y=z and x= -1, so I get (-1, z, z). We still have the y=z=0 path from before, so I get (x, 0, 0).

Using the constraint for (1, -z, z) we see that z=sqrt(3)/2 or z=-sqrt(3)/2. I get the same thing for (-1, z, z). Then for (x, 0, 0), I get x=2 and x=-2.

So in total, I think my critical points are
(1, -sqrt(3)/2, sqrt(3)/2), (1, sqrt(3)/2, -sqrt(3)/2), (-1, sqrt(3)/2, sqrt(3)/2), (-1, -sqrt(3)/2, -sqrt(3)/2), (2, 0, 0), (-2, 0, 0). I can plug those into W to find which are absolute global max and mins. And then that's basically it I believe.

And to be clear, the reason I can ignore the points inside the boundary is because the laplacian of f = 0, and if it weren't, I'd need to parameterize the f?

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BvU