Regular Equality Constraints, Multiplier Residuals and Necessity Boundaries
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batch-20261015-advance-20261119-r04-lagrange-multiplier-condition-foundations · 2026-10-15 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.
Field-order and source-binding data
1. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 1]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 1]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 1]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
2. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 2]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 2]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 2]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
3. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 3]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 3]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 3]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
4. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 4]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 4]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 4]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
5. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 5]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 5]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 5]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
6. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 6]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 6]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 6]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
7. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 7]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 7]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 7]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
8. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 8]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 8]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 8]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
9. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 9]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 9]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 9]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
10. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 10]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 10]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 10]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
11. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 11]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 11]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 11]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
12. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 12]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 12]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 12]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
13. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 13]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 13]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 13]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
14. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 14]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 14]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 14]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
15. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 15]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 15]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 15]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
16. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 16]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 16]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 16]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
17. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 17]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 17]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 17]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
18. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 18]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 18]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 18]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
19. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 19]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 19]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 19]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
20. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 20]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 20]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 20]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
21. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 21]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 21]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 21]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
22. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 22]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 22]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 22]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
23. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 23]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 23]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 23]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
24. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 24]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 24]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 24]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
25. Full Row Rank Means the Declared Constraint Gradients Are Independent
Use constraint gradients as ordered rows=[[1, 0], [0, 25]]. Give (constraint_gradient_rows, constraint_count, dimension, rank, full_row_rank, regularity_definition).
View answer and explanation
([[1, 0], [0, 25]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent)
Compute exact row rank and compare it with the number of constraints, retaining the ambient dimension. This is an algebraic qualification check on supplied vectors; it does not independently establish differentiability, feasibility or an extremum for functions not supplied here. The exact result, retaining the requested order, is ([[1, 0], [0, 25]], 2, 2, 2, Yes, full_row_rank_means_constraint_gradients_are_linearly_independent).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-regularity-audit · candidate
26. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[1, 2], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[1, 2]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([1, 2], [[1, 0], [0, 1]], [1, 2], [1, 2], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([1, 2], [[1, 0], [0, 1]], [1, 2], [1, 2], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
27. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[2, 4], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[2, 4]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([2, 4], [[1, 0], [0, 1]], [2, 4], [2, 4], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([2, 4], [[1, 0], [0, 1]], [2, 4], [2, 4], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
28. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[3, 6], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[3, 6]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([3, 6], [[1, 0], [0, 1]], [3, 6], [3, 6], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([3, 6], [[1, 0], [0, 1]], [3, 6], [3, 6], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
29. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[4, 8], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[4, 8]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([4, 8], [[1, 0], [0, 1]], [4, 8], [4, 8], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([4, 8], [[1, 0], [0, 1]], [4, 8], [4, 8], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
30. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[5, 10], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[5, 10]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([5, 10], [[1, 0], [0, 1]], [5, 10], [5, 10], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([5, 10], [[1, 0], [0, 1]], [5, 10], [5, 10], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
31. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[6, 12], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[6, 12]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([6, 12], [[1, 0], [0, 1]], [6, 12], [6, 12], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([6, 12], [[1, 0], [0, 1]], [6, 12], [6, 12], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
32. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[7, 14], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[7, 14]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([7, 14], [[1, 0], [0, 1]], [7, 14], [7, 14], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([7, 14], [[1, 0], [0, 1]], [7, 14], [7, 14], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
33. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[8, 16], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[8, 16]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([8, 16], [[1, 0], [0, 1]], [8, 16], [8, 16], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([8, 16], [[1, 0], [0, 1]], [8, 16], [8, 16], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
34. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[9, 18], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[9, 18]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([9, 18], [[1, 0], [0, 1]], [9, 18], [9, 18], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([9, 18], [[1, 0], [0, 1]], [9, 18], [9, 18], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
35. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[10, 20], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[10, 20]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([10, 20], [[1, 0], [0, 1]], [10, 20], [10, 20], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([10, 20], [[1, 0], [0, 1]], [10, 20], [10, 20], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
36. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[11, 22], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[11, 22]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([11, 22], [[1, 0], [0, 1]], [11, 22], [11, 22], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([11, 22], [[1, 0], [0, 1]], [11, 22], [11, 22], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
37. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[12, 24], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[12, 24]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([12, 24], [[1, 0], [0, 1]], [12, 24], [12, 24], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([12, 24], [[1, 0], [0, 1]], [12, 24], [12, 24], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
38. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[13, 26], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[13, 26]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([13, 26], [[1, 0], [0, 1]], [13, 26], [13, 26], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([13, 26], [[1, 0], [0, 1]], [13, 26], [13, 26], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
39. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[14, 28], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[14, 28]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([14, 28], [[1, 0], [0, 1]], [14, 28], [14, 28], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([14, 28], [[1, 0], [0, 1]], [14, 28], [14, 28], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
40. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[15, 30], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[15, 30]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([15, 30], [[1, 0], [0, 1]], [15, 30], [15, 30], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([15, 30], [[1, 0], [0, 1]], [15, 30], [15, 30], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
41. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[16, 32], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[16, 32]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([16, 32], [[1, 0], [0, 1]], [16, 32], [16, 32], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([16, 32], [[1, 0], [0, 1]], [16, 32], [16, 32], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
42. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[17, 34], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[17, 34]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([17, 34], [[1, 0], [0, 1]], [17, 34], [17, 34], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([17, 34], [[1, 0], [0, 1]], [17, 34], [17, 34], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
43. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[18, 36], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[18, 36]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([18, 36], [[1, 0], [0, 1]], [18, 36], [18, 36], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([18, 36], [[1, 0], [0, 1]], [18, 36], [18, 36], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
44. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[19, 38], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[19, 38]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([19, 38], [[1, 0], [0, 1]], [19, 38], [19, 38], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([19, 38], [[1, 0], [0, 1]], [19, 38], [19, 38], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
45. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[20, 40], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[20, 40]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([20, 40], [[1, 0], [0, 1]], [20, 40], [20, 40], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([20, 40], [[1, 0], [0, 1]], [20, 40], [20, 40], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
46. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[21, 42], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[21, 42]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([21, 42], [[1, 0], [0, 1]], [21, 42], [21, 42], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([21, 42], [[1, 0], [0, 1]], [21, 42], [21, 42], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
47. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[22, 44], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[22, 44]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([22, 44], [[1, 0], [0, 1]], [22, 44], [22, 44], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([22, 44], [[1, 0], [0, 1]], [22, 44], [22, 44], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
48. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[23, 46], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[23, 46]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([23, 46], [[1, 0], [0, 1]], [23, 46], [23, 46], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([23, 46], [[1, 0], [0, 1]], [23, 46], [23, 46], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
49. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[24, 48], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[24, 48]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([24, 48], [[1, 0], [0, 1]], [24, 48], [24, 48], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([24, 48], [[1, 0], [0, 1]], [24, 48], [24, 48], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
50. The Multiplier Equation Uses a Fixed Positive-Sign Convention
Use objective gradient=[25, 50], constraint-gradient rows=[[1, 0], [0, 1]], aligned multipliers=[25, 50]. Give (objective_gradient, constraint_gradient_rows, multipliers, gradient_combination, stationarity_residual, equation_holds, sign_convention).
This row audits declared-vector rank and the algebraic residual grad_f=J_g^T lambda only. It supplies no function/point data independently certifying feasibility, differentiability or a local extremum. equation_holds is not a full extremum certificate.
View answer and explanation
([25, 50], [[1, 0], [0, 1]], [25, 50], [25, 50], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda)
Require full row rank, compute J_g^T*lambda and residual grad_f−J_g^T*lambda in the declared coordinate order. The equation holds exactly when every residual component is0. These supplied vectors audit algebra only; separate function/point data must establish feasibility, differentiability and theorem applicability. Stationarity is not a minimum certificate. The exact result, retaining the requested order, is ([25, 50], [[1, 0], [0, 1]], [25, 50], [25, 50], [0, 0], Yes, convention_grad_f_equals_J_g_transpose_lambda).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-multiplier-residual · candidate
51. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[0, 0] subject to normal a=[1, 1], a dot x=3; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([0, 0], [1, 1], 3, 3/2, [3/2, 3/2], 3, [3, 3], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([0, 0], [1, 1], 3, 3/2, [3/2, 3/2], 3, [3, 3], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
52. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[1, 1] subject to normal a=[1, 2], a dot x=4; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([1, 1], [1, 2], 4, 1/5, [6/5, 7/5], 2/5, [2/5, 4/5], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([1, 1], [1, 2], 4, 1/5, [6/5, 7/5], 2/5, [2/5, 4/5], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
53. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[2, 2] subject to normal a=[1, 3], a dot x=5; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([2, 2], [1, 3], 5, -3/10, [17/10, 11/10], -3/5, [-3/5, -9/5], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([2, 2], [1, 3], 5, -3/10, [17/10, 11/10], -3/5, [-3/5, -9/5], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
54. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[3, 3] subject to normal a=[1, 4], a dot x=6; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([3, 3], [1, 4], 6, -9/17, [42/17, 15/17], -18/17, [-18/17, -72/17], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([3, 3], [1, 4], 6, -9/17, [42/17, 15/17], -18/17, [-18/17, -72/17], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
55. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[4, 4] subject to normal a=[1, 5], a dot x=7; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([4, 4], [1, 5], 7, -17/26, [87/26, 19/26], -17/13, [-17/13, -85/13], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([4, 4], [1, 5], 7, -17/26, [87/26, 19/26], -17/13, [-17/13, -85/13], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
56. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[0, 5] subject to normal a=[1, 6], a dot x=8; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([0, 5], [1, 6], 8, -22/37, [-22/37, 53/37], -44/37, [-44/37, -264/37], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([0, 5], [1, 6], 8, -22/37, [-22/37, 53/37], -44/37, [-44/37, -264/37], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
57. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[1, 6] subject to normal a=[1, 7], a dot x=9; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([1, 6], [1, 7], 9, -17/25, [8/25, 31/25], -34/25, [-34/25, -238/25], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([1, 6], [1, 7], 9, -17/25, [8/25, 31/25], -34/25, [-34/25, -238/25], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
58. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[2, 0] subject to normal a=[1, 8], a dot x=10; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([2, 0], [1, 8], 10, 8/65, [138/65, 64/65], 16/65, [16/65, 128/65], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([2, 0], [1, 8], 10, 8/65, [138/65, 64/65], 16/65, [16/65, 128/65], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
59. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[3, 1] subject to normal a=[1, 9], a dot x=11; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([3, 1], [1, 9], 11, -1/82, [245/82, 73/82], -1/41, [-1/41, -9/41], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([3, 1], [1, 9], 11, -1/82, [245/82, 73/82], -1/41, [-1/41, -9/41], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
60. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[4, 2] subject to normal a=[1, 10], a dot x=12; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([4, 2], [1, 10], 12, -12/101, [392/101, 82/101], -24/101, [-24/101, -240/101], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([4, 2], [1, 10], 12, -12/101, [392/101, 82/101], -24/101, [-24/101, -240/101], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
61. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[0, 3] subject to normal a=[1, 11], a dot x=13; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([0, 3], [1, 11], 13, -10/61, [-10/61, 73/61], -20/61, [-20/61, -220/61], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([0, 3], [1, 11], 13, -10/61, [-10/61, 73/61], -20/61, [-20/61, -220/61], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
62. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[1, 4] subject to normal a=[1, 12], a dot x=14; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([1, 4], [1, 12], 14, -7/29, [22/29, 32/29], -14/29, [-14/29, -168/29], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([1, 4], [1, 12], 14, -7/29, [22/29, 32/29], -14/29, [-14/29, -168/29], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
63. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[2, 5] subject to normal a=[1, 13], a dot x=15; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([2, 5], [1, 13], 15, -26/85, [144/85, 87/85], -52/85, [-52/85, -676/85], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([2, 5], [1, 13], 15, -26/85, [144/85, 87/85], -52/85, [-52/85, -676/85], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
64. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[3, 6] subject to normal a=[1, 14], a dot x=16; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([3, 6], [1, 14], 16, -71/197, [520/197, 188/197], -142/197, [-142/197, -1988/197], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([3, 6], [1, 14], 16, -71/197, [520/197, 188/197], -142/197, [-142/197, -1988/197], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
65. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[4, 0] subject to normal a=[1, 15], a dot x=17; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([4, 0], [1, 15], 17, 13/226, [917/226, 195/226], 13/113, [13/113, 195/113], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([4, 0], [1, 15], 17, 13/226, [917/226, 195/226], 13/113, [13/113, 195/113], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
66. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[0, 1] subject to normal a=[1, 16], a dot x=18; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([0, 1], [1, 16], 18, 2/257, [2/257, 289/257], 4/257, [4/257, 64/257], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([0, 1], [1, 16], 18, 2/257, [2/257, 289/257], 4/257, [4/257, 64/257], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
67. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[1, 2] subject to normal a=[1, 17], a dot x=19; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([1, 2], [1, 17], 19, -8/145, [137/145, 154/145], -16/145, [-16/145, -272/145], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([1, 2], [1, 17], 19, -8/145, [137/145, 154/145], -16/145, [-16/145, -272/145], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
68. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[2, 3] subject to normal a=[1, 18], a dot x=20; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([2, 3], [1, 18], 20, -36/325, [614/325, 327/325], -72/325, [-72/325, -1296/325], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([2, 3], [1, 18], 20, -36/325, [614/325, 327/325], -72/325, [-72/325, -1296/325], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
69. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[3, 4] subject to normal a=[1, 19], a dot x=21; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([3, 4], [1, 19], 21, -29/181, [514/181, 173/181], -58/181, [-58/181, -1102/181], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([3, 4], [1, 19], 21, -29/181, [514/181, 173/181], -58/181, [-58/181, -1102/181], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
70. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[4, 5] subject to normal a=[1, 20], a dot x=22; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([4, 5], [1, 20], 22, -82/401, [1522/401, 365/401], -164/401, [-164/401, -3280/401], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([4, 5], [1, 20], 22, -82/401, [1522/401, 365/401], -164/401, [-164/401, -3280/401], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
71. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[0, 6] subject to normal a=[1, 21], a dot x=23; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([0, 6], [1, 21], 23, -103/442, [-103/442, 489/442], -103/221, [-103/221, -2163/221], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([0, 6], [1, 21], 23, -103/442, [-103/442, 489/442], -103/221, [-103/221, -2163/221], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
72. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[1, 0] subject to normal a=[1, 22], a dot x=24; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([1, 0], [1, 22], 24, 23/485, [508/485, 506/485], 46/485, [46/485, 1012/485], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([1, 0], [1, 22], 24, 23/485, [508/485, 506/485], 46/485, [46/485, 1012/485], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
73. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[2, 1] subject to normal a=[1, 23], a dot x=25; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([2, 1], [1, 23], 25, 0, [2, 1], 0, [0, 0], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([2, 1], [1, 23], 25, 0, [2, 1], 0, [0, 0], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
74. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[3, 2] subject to normal a=[1, 24], a dot x=26; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([3, 2], [1, 24], 26, -25/577, [1706/577, 554/577], -50/577, [-50/577, -1200/577], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([3, 2], [1, 24], 26, -25/577, [1706/577, 554/577], -50/577, [-50/577, -1200/577], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
75. Squared Distance with One Regular Linear Constraint Has an Exact Solution
Minimize squared Euclidean distance to center=[4, 3] subject to normal a=[1, 25], a dot x=27; require a nonzero normal. Give (center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope).
View answer and explanation
([4, 3], [1, 25], 27, -26/313, [1226/313, 289/313], -52/313, [-52/313, -1300/313], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly)
Use displacement coefficient (rhs−a dot center)/(a dot a) along the normal. Under grad_f=J_g^T*lambda the multiplier is twice that coefficient. Verify feasibility and the complete stationarity equation. This specifically solved squared-distance/linear-constraint problem does not turn arbitrary multiplier equations into sufficient minimum, globality or uniqueness certificates. The exact result, retaining the requested order, is ([4, 3], [1, 25], 27, -26/313, [1226/313, 289/313], -52/313, [-52/313, -1300/313], Yes, Yes, squared_distance_with_one_regular_linear_constraint_solved_exactly).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-single-linear-constraint · candidate
76. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,1],[2,2]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 1], [2, 2]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 1], [2, 2]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
77. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=stationary_not_sufficient, complete model={"scale":2}. Give (mode, objective_description, constraint_description, point, stationarity_statement, stationary, actual_point_status, minimum_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(stationary_not_sufficient, f(x,y)=-2*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (stationary_not_sufficient, f(x,y)=-2*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
78. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=infeasible_point, complete model={"constraint_gradient":[1,3],"constraint_rhs":2,"point":[3,0]}. Give (mode, constraint_gradient, constraint_rhs, point, constraint_value, feasible, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(infeasible_point, [1, 3], 2, [3, 0], 3, No, stationarity_never_repairs_missing_feasibility)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (infeasible_point, [1, 3], 2, [3, 0], 3, No, stationarity_never_repairs_missing_feasibility).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
79. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=kkt_confusion, complete model={"constraint_kind":"inequality"}. Give (mode, constraint_kind, equality_theorem_supplies_KKT_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(kkt_confusion, inequality, No, equality_only_Lagrange_condition_does_not_supply_KKT_sign_or_complementarity_conditions)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (kkt_confusion, inequality, No, equality_only_Lagrange_condition_does_not_supply_KKT_sign_or_complementarity_conditions).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
80. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,5],[2,10]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 5], [2, 10]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 5], [2, 10]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
81. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=stationary_not_sufficient, complete model={"scale":6}. Give (mode, objective_description, constraint_description, point, stationarity_statement, stationary, actual_point_status, minimum_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(stationary_not_sufficient, f(x,y)=-6*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (stationary_not_sufficient, f(x,y)=-6*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
82. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=infeasible_point, complete model={"constraint_gradient":[1,7],"constraint_rhs":6,"point":[7,0]}. Give (mode, constraint_gradient, constraint_rhs, point, constraint_value, feasible, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(infeasible_point, [1, 7], 6, [7, 0], 7, No, stationarity_never_repairs_missing_feasibility)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (infeasible_point, [1, 7], 6, [7, 0], 7, No, stationarity_never_repairs_missing_feasibility).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
83. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,8],[2,16]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 8], [2, 16]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 8], [2, 16]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
84. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,9],[2,18]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 9], [2, 18]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 9], [2, 18]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
85. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=stationary_not_sufficient, complete model={"scale":10}. Give (mode, objective_description, constraint_description, point, stationarity_statement, stationary, actual_point_status, minimum_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(stationary_not_sufficient, f(x,y)=-10*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (stationary_not_sufficient, f(x,y)=-10*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
86. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=infeasible_point, complete model={"constraint_gradient":[1,11],"constraint_rhs":10,"point":[11,0]}. Give (mode, constraint_gradient, constraint_rhs, point, constraint_value, feasible, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(infeasible_point, [1, 11], 10, [11, 0], 11, No, stationarity_never_repairs_missing_feasibility)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (infeasible_point, [1, 11], 10, [11, 0], 11, No, stationarity_never_repairs_missing_feasibility).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
87. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,12],[2,24]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 12], [2, 24]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 12], [2, 24]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
88. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,13],[2,26]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 13], [2, 26]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 13], [2, 26]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
89. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=stationary_not_sufficient, complete model={"scale":14}. Give (mode, objective_description, constraint_description, point, stationarity_statement, stationary, actual_point_status, minimum_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(stationary_not_sufficient, f(x,y)=-14*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (stationary_not_sufficient, f(x,y)=-14*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
90. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=infeasible_point, complete model={"constraint_gradient":[1,15],"constraint_rhs":14,"point":[15,0]}. Give (mode, constraint_gradient, constraint_rhs, point, constraint_value, feasible, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(infeasible_point, [1, 15], 14, [15, 0], 15, No, stationarity_never_repairs_missing_feasibility)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (infeasible_point, [1, 15], 14, [15, 0], 15, No, stationarity_never_repairs_missing_feasibility).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
91. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,16],[2,32]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 16], [2, 32]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 16], [2, 32]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
92. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,17],[2,34]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 17], [2, 34]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 17], [2, 34]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
93. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=stationary_not_sufficient, complete model={"scale":18}. Give (mode, objective_description, constraint_description, point, stationarity_statement, stationary, actual_point_status, minimum_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(stationary_not_sufficient, f(x,y)=-18*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (stationary_not_sufficient, f(x,y)=-18*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
94. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=infeasible_point, complete model={"constraint_gradient":[1,19],"constraint_rhs":18,"point":[19,0]}. Give (mode, constraint_gradient, constraint_rhs, point, constraint_value, feasible, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(infeasible_point, [1, 19], 18, [19, 0], 19, No, stationarity_never_repairs_missing_feasibility)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (infeasible_point, [1, 19], 18, [19, 0], 19, No, stationarity_never_repairs_missing_feasibility).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
95. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,20],[2,40]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 20], [2, 40]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 20], [2, 40]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
96. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,21],[2,42]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 21], [2, 42]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 21], [2, 42]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
97. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=stationary_not_sufficient, complete model={"scale":22}. Give (mode, objective_description, constraint_description, point, stationarity_statement, stationary, actual_point_status, minimum_claim, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(stationary_not_sufficient, f(x,y)=-22*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (stationary_not_sufficient, f(x,y)=-22*x^2, g(x,y)=y=0, [0,0], grad_f=[0,0]=0*grad_g, Yes, strict_local_maximum_on_constraint, No, first_order_multiplier_equation_is_necessary_not_sufficient_for_a_minimum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
98. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=infeasible_point, complete model={"constraint_gradient":[1,23],"constraint_rhs":22,"point":[23,0]}. Give (mode, constraint_gradient, constraint_rhs, point, constraint_value, feasible, boundary).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(infeasible_point, [1, 23], 22, [23, 0], 23, No, stationarity_never_repairs_missing_feasibility)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (infeasible_point, [1, 23], 22, [23, 0], 23, No, stationarity_never_repairs_missing_feasibility).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
99. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,24],[2,48]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 24], [2, 48]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 24], [2, 48]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
100. Regular Equality Conditions Are Necessary, Not Sufficient, and Not Inequality KKT
Use mode=rank_deficient, complete model={"constraint_gradients":[[1,25],[2,50]]}. Give (mode, constraint_gradient_rows, rank, constraint_count, regular_theorem_applicable, scope).
The four modes have6/9/7/4 fields. Rank deficiency withholds this regular theorem; it does not imply no extremum. Necessary stationarity is not sufficient and does not repair feasibility or supply inequality KKT conditions.
View answer and explanation
(rank_deficient, [[1, 25], [2, 50]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum)
Dependent constraint gradients withhold this regular theorem but do not imply absence of extrema. A feasible stationary point can be a strict constrained maximum. Stationarity never repairs infeasibility. Equality-only multipliers do not supply inequality KKT signs or complementarity, nor general global or unique primal optimality. The exact result, retaining the requested order, is (rank_deficient, [[1, 25], [2, 50]], 1, 2, No, regular_Lagrange_theorem_not_applicable_does_not_mean_no_extremum).
Reviewed step-by-step derivation
Necessary Lagrange multiplier condition for equality constraints · qf-mod50-lagrange-boundary · candidate
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