bottom-left: same as upper-right, but with a contour representation, which shows more clearly the contour lines and the extremums of the three functions involved.When equilibrium is not forced, it is a function of the three variables,. When equilibrium is forced, its extremum corresponds to the solution of the problem. The function plotted in green is the new potential, which is to be optimized without constraint on. The constraint function is rescaled ( ), but as, the constraint (thick blue line) keeps the same position as in the previous case.

upper-right: this is the case treated with the help of.The thick blue line is the constraint, the thick red line is its projection on, and the solution is the top of the red thick line.

upper-left: this is the case treated without the Lagrange multiplier.

Here are a few explanations for each of the four plots displayed: Is then a tuning tool to find the place where the functions and have the same slope (strict parallelism) and hence where the potential is an extremum (equilibrium situation). on a contour line or (the constant is unknown, as the contour line is what we are looking for) tangent to the previous one.īy imposing (for any ), we impose that and share the same orientation but with different relative slopes (depending on ). The global problem can be understood as finding a point that is:Ģ. This in turn shows how the independence of and is restored by the introduction of (which is adjusted so that and are strictly parallel in both slope and orientation, so that their height difference remains constant near the solution, with no more constraint between and ). This Demonstration illustrates the 2D case, where in particular, the Lagrange multiplier is shown to modify not only the relative slopes of the function to be minimized and the rescaled constraint (which was already shown in the 1D case), but also their relative orientations (which do not exist in the 1D case). See Lagrange Multipliers in One Dimension. The main ideas behind the Lagrange multipliers have already been discussed in the 1D case. The equilibrium condition ( ) can be seen as a constraint on. The function (plotted in green) is the difference in height between and the rescaled constraint function, which is constant only when and are parallel.

$mathematica 11.3 plot sphere$

Hence, as, for obvious symmetry reasons the complete solution here is quite simple. In this particular case, the new potential is a Legendre transformation, where (given the constraint ). Here, so the constraint is, which is simpler to visualize (thick blue line). The constraint is its particular contour line. The function is the constraint function, plotted in blue. The function (plotted in red) is the one to be optimized subject to the constraint.

Mathematica 11.3 plot sphere