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ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A constrained extremum on an affine graph satisfies the graph Lagrange rule

Statement

The minimum of f(x,y)=x2+y2f(x,y)=x^2+y^2 on the graph y=x+1y=x+1 occurs at (−1/2,1/2)(-1/2,1/2) and satisfies ∇f=λ∇(y−x−1)\nabla f=\lambda\nabla(y-x-1) with λ=1\lambda=1.

Facts & Assumptions

Given: f(x,y)=x2+y2f(x,y)=x^2+y^2 and G(x,y)=y−x−1G(x,y)=y-x-1.

[L1]

The graph-constraint Lagrange rule is Lagrange multipliers for a regular graph constraint y=ψ(x)y=\psi(x).

Proof

technique · calculation
1.1

On the graph, f(x,x+1)=2x2+2x+1=2(x+1/2)2+1/2f(x,x+1)=2x^2+2x+1=2(x+1/2)^2+1/2, so the unique constrained minimum is a=(−1/2,1/2)a=(-1/2,1/2).

givenalgebra
1.2

At aa, ∇f(a)=(−1,1)=∇G(a)\nabla f(a)=(-1,1)=\nabla G(a). Hence the multiplier equation holds with λ=1\lambda=1, in accord with [L1].

givenL1algebra
2.1

This explicitly realizes the graph-constraint conclusion at the constrained minimum.

step 1.1step 1.2algebra∎

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