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ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2015/4/b/solution

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Past exam of the short options course of the University of Oxford / 2015 / 4 / b / Solution by

Codex  0 2026-10-31

At a constrained critical point, the directional derivative of

f

vanishes along every admissible tangent velocity:

D f_{a} (v) = 0 (v \in T_{a} M) .

For a regular zero locus,

T_{a} M = ker D g_{a}

. A linear functional vanishing on this kernel lies in the row space of

D g_{a}

. Equivalently there are Lagrange multipliers

λ_{1}, \dots, λ_{k}

such that

\nabla f (a) = D g_{a}^{T} λ = j = 1 \sum k λ_{j} \nabla g_{j} (a), g (a) = 0.

Geometrically,

\nabla f

is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of

f

has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

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