Re: Lagrange Multiplier Problem, three variables

JTolman wrote:
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


Sorry about that, I forgot that you were solving w/ the Lagrange multiplier.

f[ x_,y_,z_]:= 1/ x^2+ 1/ y^2+ 1/ z^2
g[ x_,y_,z_]:= x y z-v
sols= Reduce[ D[ f[ x,y,z], { { x,y,z}}]== \[Lambda] D[ g[ x,y,z], { { x,y,z}}]&& x>0&& y>0&& z>0&& x y z==v&& v>0, { x,y,z,\[Lambda]}]

returns

v > 0 && x == Root[-v + #1^3 & , 1] && y == v/x^2 && z == v/(x*y) && \[Lambda] == -(2/(x*y*z^3))

You can just "discard" \[Lambda] since x, y, and z are independent of it.

HTH,
Jean-Marc
.

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