Lagrange Multipliers to find max/min values

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Homework Help Overview

The discussion revolves around using Lagrange multipliers to find the maximum and minimum values of the function f(x,y,z) = x²y²z², subject to the constraint x² + y² + z² = 1. Participants are exploring the implications of this method and the relationships between the variables under the given constraint.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to apply the method of Lagrange multipliers, expressing the gradients and setting up equations based on the relationships between the variables and the multiplier. There is confusion regarding the values of x, y, and z, particularly when assuming they are equal and how that fits within the constraint. Some participants are questioning the validity of their assumptions and the implications of their findings.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and reasoning. Some have provided insights into potential solutions and the nature of stationary points, while others express uncertainty about their conclusions and the correctness of their approaches. There is no explicit consensus, but various interpretations and methods are being explored.

Contextual Notes

Participants are assuming that x, y, z > 0, and there is mention of the possibility of other cases being considered later. The constraint x² + y² + z² = 1 is central to the discussion, and participants are grappling with how to satisfy this while exploring different values for x, y, and z.

arl146
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Homework Statement


Use Lagrange multipliers to find the max and min values of the function subject to the given constraints:

f(x,y,z)= x2y2z2
constraint: x2 + y2 + z2 = 1


Homework Equations


∇f = ∇g * λ
fx = gx * λ
fy = gy * λ
fz = gz * λ

The Attempt at a Solution


i can't solve for x, y, and lambda

i got:
for this one, fx = gx * λ ---> 2x*y2*z2=(2x)λ
for, fy = gy * λ ---> 2y*x2*z2=(2y)λ
fz = gz * λ ----> 2z*x2*y2=(2z)λ

i tried setting them all equal to lambda, and ended up getting that x=y=z=1 but when you add that into g(x,y,z)... that equals >1 ? I am confused with getting the x,y,z, and lambda values. help?
 
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oh please help!
 
arl146 said:

Homework Statement


Use Lagrange multipliers to find the max and min values of the function subject to the given constraints:

f(x,y,z)= x2y2z2
constraint: x2 + y2 + z2 = 1


Homework Equations


∇f = ∇g * λ
fx = gx * λ
fy = gy * λ
fz = gz * λ

The Attempt at a Solution


i can't solve for x, y, and lambda

i got:
for this one, fx = gx * λ ---> 2x*y2*z2=(2x)λ
for, fy = gy * λ ---> 2y*x2*z2=(2y)λ
fz = gz * λ ----> 2z*x2*y2=(2z)λ

i tried setting them all equal to lambda, and ended up getting that x=y=z=1 but when you add that into g(x,y,z)... that equals >1 ? I am confused with getting the x,y,z, and lambda values. help?


Assuming that x, y, z > 0 (other cases later) the first equation 2x*y^2*z^2= (2x)*u [using u instead of lambda] can be divided through by 2x on both sides to get y^2*z^2=u, that is: (yz)^2 = u. Similarly, (xz)^2 = u and (xy)^2 = u, so xz = xy = yz = v, v = sqrt(u) (since we have assumed all x, y, z > 0). So we get x = y = z. Now use the constraint.

Note that there are lots of other stationary points, essentially differing by signs in each of x, y and z. Also: you need to check whether you can have solutions with, for example, x = 0.


RGV
 
I think I'm working on something similar... this is what I would have done

[tex]\begin{gathered}<br /> f\left( {x,y,z} \right) = {x^2}{y^2}{z^2};g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 1 = 0 \\<br /> \vec \nabla f = \left\langle {2x{y^2}{z^2},2{x^2}y{z^2},2{x^2}{y^2}z} \right\rangle ;\vec \nabla g = \left\langle {2x,2y,2z} \right\rangle \\<br /> \vec \nabla f = \lambda \vec \nabla g \to {y^2}{z^2} = {x^2}{z^2} = {x^2}{y^2} \to x = \pm y = \pm z \\<br /> 3{x^2} = 1 \to {x^2} = \frac{1}{3} \to x = \pm \frac{{\sqrt 3 }}{3} \\<br /> f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = \\<br /> f\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3} - \frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = \frac{1}{{27}} \\ <br /> \end{gathered}[/tex]
 
Ray Vickson said:
Assuming that x, y, z > 0 (other cases later) the first equation 2x*y^2*z^2= (2x)*u [using u instead of lambda] can be divided through by 2x on both sides to get y^2*z^2=u, that is: (yz)^2 = u. Similarly, (xz)^2 = u and (xy)^2 = u, so xz = xy = yz = v, v = sqrt(u) (since we have assumed all x, y, z > 0). So we get x = y = z. Now use the constraint.

Note that there are lots of other stationary points, essentially differing by signs in each of x, y and z. Also: you need to check whether you can have solutions with, for example, x = 0.


RGV

i did that whole thing... i got that x=y=z and so I am confused because how does x^2 + y^2 + z^2 = 1 if x=y=z ?
 
arcyqwerty: didnt even think about doing that way the 3x^2=1
 
so is arcyqwerty right? haha is there really those 8 points and only those?
 
Last edited by a moderator:
  • #10
going off this, but using it for another problem .. i think i see a trend...
whatever 'n' is, is what the values are. like how we found that x=y=z= 1/sqrt(3) so not using values, just using 'n', we can assume that each value will always equal 1/sqrt(n) with just these types of problems... where you have f(x1, x2, ... , xn)= x1+x2+...+xn and the constraint is x1^2 + x2^2 + ... + xn^2
 

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