Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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+y2 on the graph y=x+1 occurs at (−1/2,1/2) and satisfies ∇f=λ∇(y−x−1) with λ=1.

Facts & Assumptions

Given: f(x,y)=x2+y2 and G(x,y)=y−x−1.

[L1]

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

Proof

technique · calculation
1.1

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

givenalgebra
1.2

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

givenL1algebra
2.1

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

step 1.1step 1.2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources