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 August 6th, 2017, 05:36 PM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics Solve feature mathematica I have been using the Solve feature in mathematica to solve the system of equations for me when using Lagrange Multipliers. It has been a great way to check my work, but now I just began adding another Lagrange multiplier, and now mathematica is not giving me anything for an output. Perhaps I am missing something small here? Solve[{ x^2 + y^2 - 2 == 0, x + z - 1 == 0, Grad[x + y + z, {x, y, z}] == L1 Grad[x^2 + y^2 - 2, {x, y}] + L2 Grad[x + z - 1, {x, z}] }] I also tried Solve[{ x^2 + y^2 - 2 == 0, x + z - 1 == 0, Grad[x + y + z, {x, y, z}] == L1 Grad[x^2 + y^2 - 2, {x, y,z}] + L2 Grad[x + z - 1, {x,y, z}] }] and still no cigar. Any ideas? Huge thanks! Jacob
 August 6th, 2017, 09:27 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 you need to include the multipliers L1 and L2 in the solve for list. Mathematica is pretty picky about wanting the equation list and the solve for list to have the same length. Thanks from SenatorArmstrong
 August 8th, 2017, 10:04 PM #3 Newbie   Joined: Aug 2017 From: Vancouver Posts: 1 Thanks: 0 Code: Simplify[{ x^2 + y^2 - 2 == 0, x + z - 1 == 0, Grad[x + y + z, {x, y, z}] == L1 Grad[x^2 + y^2 - 2, {x, y}] + L2 Grad[x + z - 1, {x, z}] }] shows the third equation is False, so it thinks there can be no solution to that. Code: Reduce[{ x^2 + y^2 - 2 == 0, x + z - 1 == 0, Grad[x + y + z, {x, y, z}] == L1 Grad[x^2 + y^2 - 2, {x, y, z}] + L2 Grad[x + z - 1, {x, y, z}]}, {x, y, z}] but not Solve, does find L1, L2, x, y, z to satisfy that. Notice that I did include the list of variables for it to find.

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