第 28000 題：Lagrange乘子條件

原題與已知條件

feasibility、constraint-gradient full row rank與grad_f=J_g^T*lambda convention。操作=lagrange-single-linear-constraint；明示資料={"center":[4,3],"constraint_gradient":[1,25],"constraint_rhs":27}；請依 exact_ordered_tuple_declared_by_interface 回答。

• G001 — feasibility、constraint-gradient full row rank與grad_f=J_g^T*lambda convention。操作=lagrange-single-linear-constraint；明示資料={"center":[4,3],"constraint_gradient":[1,25],"constraint_rhs":27}；請依 exact_ordered_tuple_declared_by_interface 回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-lagrange-multiplier-condition

逐步推導

source roles = [lagrange-single-linear-constraint]

derived = [[center, constraint_gradient, constraint_rhs, normal_displacement_coefficient, solution_point, multiplier, objective_gradient, feasible, stationary, specified_problem_scope]]

Lagrange single ORIGINAL normal squared norm ORIGINAL product0 := mul(1, 1)

Lagrange single ORIGINAL normal squared norm ORIGINAL product0 = 1

Lagrange single ORIGINAL normal squared norm ORIGINAL running sum0 := sum(0, 1)

Lagrange single ORIGINAL normal squared norm ORIGINAL running sum0 = 1

Lagrange single ORIGINAL normal squared norm ORIGINAL product1 := mul(25, 25)

Lagrange single ORIGINAL normal squared norm ORIGINAL product1 = 625

Lagrange single ORIGINAL normal squared norm ORIGINAL running sum1 := sum(1, 625)

Lagrange single ORIGINAL normal squared norm ORIGINAL running sum1 = 626

Lagrange single ORIGINAL normal dot center ORIGINAL product0 := mul(1, 4)

Lagrange single ORIGINAL normal dot center ORIGINAL product0 = 4

Lagrange single ORIGINAL normal dot center ORIGINAL running sum0 := sum(0, 4)

Lagrange single ORIGINAL normal dot center ORIGINAL running sum0 = 4

Lagrange single ORIGINAL normal dot center ORIGINAL product1 := mul(25, 3)

Lagrange single ORIGINAL normal dot center ORIGINAL product1 = 75

Lagrange single ORIGINAL normal dot center ORIGINAL running sum1 := sum(4, 75)

Lagrange single ORIGINAL normal dot center ORIGINAL running sum1 = 79

Lagrange single ORIGINAL rhs minus center value := sub(27, 79)

Lagrange single ORIGINAL rhs minus center value = -52

Lagrange single ORIGINAL normal displacement := div(-52, 626)

Lagrange single ORIGINAL normal displacement = -26/313

Lagrange single ORIGINAL normal shift0 := mul(-26/313, 1)

Lagrange single ORIGINAL normal shift0 = -26/313

Lagrange single ORIGINAL solution coordinate0 := sum(4, -26/313)

Lagrange single ORIGINAL solution coordinate0 = 1226/313

Lagrange single ORIGINAL normal shift1 := mul(-26/313, 25)

Lagrange single ORIGINAL normal shift1 = -650/313

Lagrange single ORIGINAL solution coordinate1 := sum(3, -650/313)

Lagrange single ORIGINAL solution coordinate1 = 289/313

Lagrange single ORIGINAL multiplier2tau := mul(2, -26/313)

Lagrange single ORIGINAL multiplier2tau = -52/313

Lagrange single ORIGINAL displacement coordinate0 := sub(1226/313, 4)

Lagrange single ORIGINAL displacement coordinate0 = -26/313

Lagrange single ORIGINAL objective gradient0 := mul(2, -26/313)

Lagrange single ORIGINAL objective gradient0 = -52/313

Lagrange single ORIGINAL displacement coordinate1 := sub(289/313, 3)

Lagrange single ORIGINAL displacement coordinate1 = -650/313

Lagrange single ORIGINAL objective gradient1 := mul(2, -650/313)

Lagrange single ORIGINAL objective gradient1 = -1300/313

Lagrange single ORIGINAL full feasibility ORIGINAL product0 := mul(1, 1226/313)

Lagrange single ORIGINAL full feasibility ORIGINAL product0 = 1226/313

Lagrange single ORIGINAL full feasibility ORIGINAL running sum0 := sum(0, 1226/313)

Lagrange single ORIGINAL full feasibility ORIGINAL running sum0 = 1226/313

Lagrange single ORIGINAL full feasibility ORIGINAL product1 := mul(25, 289/313)

Lagrange single ORIGINAL full feasibility ORIGINAL product1 = 7225/313

Lagrange single ORIGINAL full feasibility ORIGINAL running sum1 := sum(1226/313, 7225/313)

Lagrange single ORIGINAL full feasibility ORIGINAL running sum1 = 27

Lagrange single ORIGINAL lambda times normal0 := mul(-52/313, 1)

Lagrange single ORIGINAL lambda times normal0 = -52/313

Lagrange single ORIGINAL lambda times normal1 := mul(-52/313, 25)

Lagrange single ORIGINAL lambda times normal1 = -1300/313

source roles = [[4, 3], [1, 25], 27, 626, 79, -26/313, [1226/313, 289/313], -52/313, [-52/313, -1300/313]]

derived = [是, 是, ANY feasible y=x+v has a dot v=0;f(y)=f(x)+norm(v)^2]

逐步反向核對

• S027 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S026 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL lambda times normal1，不以小數近似替代。
• S025 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL lambda times normal0，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL full feasibility ORIGINAL running sum1，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL full feasibility ORIGINAL product1，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL full feasibility ORIGINAL running sum0，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL full feasibility ORIGINAL product0，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL objective gradient1，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL displacement coordinate1，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL objective gradient0，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL displacement coordinate0，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL multiplier2tau，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL solution coordinate1，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal shift1，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL solution coordinate0，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal shift0，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal displacement，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL rhs minus center value，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal dot center ORIGINAL running sum1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal dot center ORIGINAL product1，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal dot center ORIGINAL running sum0，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal dot center ORIGINAL product0，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal squared norm ORIGINAL running sum1，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal squared norm ORIGINAL product1，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal squared norm ORIGINAL running sum0，不以小數近似替代。
• S002 · premise_and_result_audit — 以精確有理數重算 Lagrange single ORIGINAL normal squared norm ORIGINAL product0，不以小數近似替代。
• S001 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。

結論

([4, 3], [1, 25], 27, -26/313, [1226/313, 289/313], -52/313, [-52/313, -1300/313], 是, 是, squared_distance_with_one_regular_linear_constraint_solved_exactly)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。