Lagrange Multipliers to find max/min values

[arl146]

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
f_x = g_x * λ
f_y = g_y * λ
f_z = g_z * λ

The Attempt at a Solution

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

i got:
for this one, f_x = g_x * λ ---> 2x*y²*z²=(2x)λ
for, f_y = g_y * λ ---> 2y*x²*z²=(2y)λ
f_z = g_z * 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?

 

[arl146]

oh please help!

 

[Ray Vickson]

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
f_x = g_x * λ
f_y = g_y * λ
f_z = g_z * λ

The Attempt at a Solution

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

i got:
for this one, f_x = g_x * λ ---> 2x*y²*z²=(2x)λ
for, f_y = g_y * λ ---> 2y*x²*z²=(2y)λ
f_z = g_z * 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

 

[arcyqwerty]

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

$$\begin{gathered} f\left( {x,y,z} \right) = {x^2}{y^2}{z^2};g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 1 = 0 \\ \vec \nabla f = \left\langle {2x{y^2}{z^2},2{x^2}y{z^2},2{x^2}{y^2}z} \right\rangle ;\vec \nabla g = \left\langle {2x,2y,2z} \right\rangle \\ \vec \nabla f = \lambda \vec \nabla g \to {y^2}{z^2} = {x^2}{z^2} = {x^2}{y^2} \to x = \pm y = \pm z \\ 3{x^2} = 1 \to {x^2} = \frac{1}{3} \to x = \pm \frac{{\sqrt 3 }}{3} \\ f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = \\ f\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3} - \frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = \frac{1}{{27}} \\ \end{gathered}$$

 

[arl146]

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 ?

 

[arl146]

arcyqwerty: didnt even think about doing that way the 3x^2=1

 

[arl146]

so is arcyqwerty right? haha is there really those 8 points and only those?

 

[arcyqwerty]

so is arcyqwerty right? haha is there really those 8 points and only those?

Well, that's as far as I got with my similar problem https://www.physicsforums.com/showthread.php?p=3582731&posted=1#post3582731"

That might just be a max or something. Not entirely sure.

 

[arl146]

Well, that's as far as I got with my similar problem https://www.physicsforums.com/showthread.php?p=3582731&posted=1#post3582731"

That might just be a max or something. Not entirely sure.

i mean it TOTALLY makes sense... but i am the one here for help, so idk if i trust myself haha

 

[arl146]

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, ... , x_n)= x₁+x₂+...+x_n and the constraint is x₁^2 + x₂^2 + ... + x_n^2