Re: Lagrange Multiplier Problem, three variables

JTolman schrieb:
Jean-Marc Gulliet wrote:
Assigning a value to lambda might help?

eqn1= 2/ x^3== \[Lambda] y z;
eqn2= 2/ y^3 == \[Lambda] x z;
eqn3= 2/ z^3== \[Lambda] x y;
N@ Solve[ { eqn1,eqn2,eqn3}, { x,y,z}]/. \[Lambda]->1

For the numbers I tried I got complex numbers, which I am fairly
certain wouldn't be valid for the application problem I am working on.

Also, you could try Reduce:

--> ((x == -1.1486983549970349 && (y == x || y == -1.*x)) || (x ==
1.148698354997035 && (y == -1.*x || y == x))) && z == 2./(x*y^3)

Am I interpreting this right, (-1.14869, x, 2/xy^3, (-1.14869, -x, 2/x
y ^3), (1.14869, x, 2/x y ^3), (1.14869, -x, 2/x y ^3)

And this is the solution for lambda = 1, but I thought the way lagrange
multipliers worked you had to actually solve for lambda, you couldn't
just plug any number in?
HTH,
Jean-Marc


It is easy, to solve the equations w.r.t. x,y,z and lambda and select the meaningful {x, y, z} - triple(s):

In[1]:=
Select[{x, y, z} /.
Solve[D[1/x^2 + 1/y^2 + 1/z^2 + lambda*(v - x*y*z), {{x, y, z}}] == 0
&& v == x*y*z, {x, y, z, lambda}],
Refine[And @@ (# > 0 & /@ #, v > 0] & ]

Out[1]=
{{v^(1/3), v^(1/3), v^(1/3)}}

But it doesn't need a single line of Mathematica-code to argue with symmetry and the singularities at x|y|z = 0, that the solution has x=y=z and therefore x^3==v and your min = 3*v^(-2/3).

hth,
Peter
.

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