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

∇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?

arl146

Homework Helper
Dearly Missed

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

∇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

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?

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