Re: Lagrange Multiplier Problem, three variables


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

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