derivation-mko-lagrange-multiplier-condition-01 · ocme-new142-reader-adoption-20261021
Minimize the squared distance to (1,2) subject to x_1+x_2=0; find the point and multiplier and independently prove the minimum.
Fix the positive sign grad f=J_g^T*lambda; general multipliers are necessary only, while this particular distance problem has a separate sufficiency proof.
algebrapolynomiallagrange_regularG1 — f(x_1,x_2)=(x_1-1)^2+(x_2-2)^2, g=x_1+x_2; the Jacobian rows follow the single constraint's order.J_g — row Jacobian ordered by constraint\lambda — multiplier in the positive-sign conventionv — arbitrary feasible differencemko_definition · cited — Source semantic paraphrase: at a feasible local extremum, neighborhood differentiability and full row rank give a unique multiplier representing the objective gradient; the condition is necessary only. · mko-lagrange-multiplier-conditionalgebra · prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.polynomial · prerequisite — Real polynomials are continuous and differentiate termwise: the derivative of x^k is k*x^(k-1), and constants differentiate to zero.lagrange_regular · prerequisite — At a feasible local extremum, neighborhood differentiability and independent constraint-gradient rows imply a unique lambda with grad f=J_g^T*lambda; this is necessary only.Before
n=2,\quad m=1,\quad J_g=(1\ \ 1)
After
\operatorname{rank}J_g=1=m,\quad f,g\text{ differentiable on }\mathbb R^2Conditions: none
Rule: polynomial · Uses: G1
Leaf basis: polynomial
The single nonzero row has full row rank, and the full polynomials give neighborhood differentiability.
Before
\nabla f=(2(x_1-1),2(x_2-2))
After
2(x_1-1)=\lambda,\quad2(x_2-2)=\lambda,\quad x_1+x_2=0
Conditions: none
Rule: lagrange_regular · Uses: G1, S01, lagrange_regular
Leaf basis: lagrange_regular
Form the candidate equations from necessary stationarity and feasibility without yet declaring their solution a minimum.
Before
x_1=1+\lambda/2,\quad x_2=2+\lambda/2
After
3+\lambda=0\Longrightarrow\lambda=-3
Conditions: none
Rule: algebra · Uses: S02
Leaf basis: algebra
Substitution into the single linear constraint determines the multiplier uniquely.
Before
\lambda=-3
After
x^*=(-1/2,1/2)
Conditions: none
Rule: algebra · Uses: S03
Leaf basis: algebra
Substitute into both candidate-coordinate formulas.
Before
g(x^*)=-1/2+1/2
After
g(x^*)=0,\quad\nabla f(x^*)=(-3,-3)=J_g^T(-3)
Conditions: none
Rule: algebra · Uses: G1, S01, S04
Leaf basis: algebra
Independently recheck feasibility and stationarity residual at the candidate; full rank remains valid.
Before
z=(1,2),\quad x^*-z=(-3/2,-3/2)
After
f(x^*)=9/4+9/4=9/2
Conditions: none
Rule: algebra · Uses: G1, S04, S05
Leaf basis: algebra
Compute the squared distance exactly as the reference for all comparison points.
Before
y=x^*+v\text{ feasible}\Longrightarrow v_1+v_2=0After
f(y)=\|x^*-z\|^2+2(x^*-z)\cdot v+\|v\|^2=9/2+\|v\|^2\ge9/2
Conditions: y\text{ feasible}
Rule: algebra · Uses: G1, S04, S06
Leaf basis: algebra
The offset x*-z is parallel to normal (1,1), orthogonal to every feasible difference v, so the cross term vanishes. Equality forces v=0, separately proving the unique global minimizer.
Before
F(u,v)=u^3,\quad G(u,v)=v,\quad(u,v)=(0,0)
After
\nabla F=0=J_G^T\cdot0,\quad\operatorname{rank}J_G=1;\quad F(t,0)=t^3\text{ has both signs}Conditions: none
Rule: polynomial · Uses: G1
Leaf basis: polynomial
A regular feasible stationary point can still fail to be an extremum, explicitly showing necessity is not sufficiency.
No substeps in this guide.
S08 · premise_audit · G1 — The main distance example's minimum was independently proved at S07 and is not conflated with this counterexample.S07 · premise_audit · G1, S04, S06 — Sufficiency and globality come from this squared decomposition, not the general necessary multiplier theorem.S06 · premise_audit · G1, S04, S05 — The numerical value alone does not prove minimality.S05 · premise_audit · G1, S01, S04 — The constraint is not omitted after solving for lambda.S04 · inverse_equivalence · S03 — The center remains (1,2), not the origin.S03 · inverse_equivalence · S02 — A unique multiplier does not imply a unique primal point for general problems.S02 · premise_audit · G1, S01, lagrange_regular — The positive-sign convention is fixed, not silently changed to grad f+lambda*grad g=0.S01 · premise_audit · G1 — Both m≤n and row/column conventions are checked.Check feasibility, neighborhood differentiability and full row rank before applying the necessary theorem.
The global minimum follows from a squared decomposition for every feasible y, not lambda alone.
reject — The regular stationary example at S08 is not an extremum.reject — In one dimension g(x)=x^2=0 has only feasible point zero, where f(x)=x attains an extremum; g'(0)=0 only withholds the theorem.reject — Inequality multiplier signs and complementary slackness are absent.x*=(-1/2,1/2), lambda=-3 and minimum squared distance 9/2; uniqueness and globality are separately proved by orthogonal squared decomposition.
2 declared checks for this guide; only the listed expressions are covered.
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