第 25079 題：線性規劃標準形

原題與已知條件

固定maximize cTx、Ax≤b、x≥0的唯一standard-form convention。操作=lp-standard-unrestricted-split；資料={"column":[4,-5,6],"objective_coefficient":7,"value":-4}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 固定maximize cTx、Ax≤b、x≥0的唯一standard-form convention。操作=lp-standard-unrestricted-split；資料={"column":[4,-5,6],"objective_coefficient":7,"value":-4}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

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-linear-program-standard-form

逐步推導

source roles = [lp-standard-unrestricted-split]

derived = [[column, objective_coefficient, value, x_plus, x_minus, reconstructed_value, original_column_contribution, split_column_contribution, columns_equal, original_objective_contribution, split_objective_contribution, objectives_equal, certificate]]

LP unrestricted reconstruct value := sub(0, 4)

LP unrestricted reconstruct value = -4

LP unrestricted ORIGINAL column0 := mul(4, -4)

LP unrestricted ORIGINAL column0 = -16

LP unrestricted positive column0 := mul(4, 0)

LP unrestricted positive column0 = 0

LP unrestricted negative column0 := mul(-4, 4)

LP unrestricted negative column0 = -16

LP unrestricted split column0 := sum(0, -16)

LP unrestricted split column0 = -16

LP unrestricted ORIGINAL column1 := mul(-5, -4)

LP unrestricted ORIGINAL column1 = 20

LP unrestricted positive column1 := mul(-5, 0)

LP unrestricted positive column1 = 0

LP unrestricted negative column1 := mul(5, 4)

LP unrestricted negative column1 = 20

LP unrestricted split column1 := sum(0, 20)

LP unrestricted split column1 = 20

LP unrestricted ORIGINAL column2 := mul(6, -4)

LP unrestricted ORIGINAL column2 = -24

LP unrestricted positive column2 := mul(6, 0)

LP unrestricted positive column2 = 0

LP unrestricted negative column2 := mul(-6, 4)

LP unrestricted negative column2 = -24

LP unrestricted split column2 := sum(0, -24)

LP unrestricted split column2 = -24

LP unrestricted ORIGINAL objective contribution := mul(7, -4)

LP unrestricted ORIGINAL objective contribution = -28

LP unrestricted positive objective := mul(7, 0)

LP unrestricted positive objective = 0

LP unrestricted negative objective := mul(-7, 4)

LP unrestricted negative objective = -28

LP unrestricted split objective sum := sum(0, -28)

LP unrestricted split objective sum = -28

source roles = [[4, -5, 6], 7, -4, 0, 4]

derived = [-4, [-16, 20, -24], [-16, 20, -24], 是, -28, -28, 是]

逐步反向核對

• S019 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S018 · premise_and_result_audit — 以精確有理數重算 LP unrestricted split objective sum，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 LP unrestricted negative objective，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 LP unrestricted positive objective，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 LP unrestricted ORIGINAL objective contribution，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 LP unrestricted split column2，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 LP unrestricted negative column2，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 LP unrestricted positive column2，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 LP unrestricted ORIGINAL column2，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 LP unrestricted split column1，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 LP unrestricted negative column1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 LP unrestricted positive column1，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 LP unrestricted ORIGINAL column1，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 LP unrestricted split column0，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 LP unrestricted negative column0，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 LP unrestricted positive column0，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 LP unrestricted ORIGINAL column0，不以小數近似替代。
• S002 · premise_and_result_audit — 以精確有理數重算 LP unrestricted reconstruct value，不以小數近似替代。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

([4, -5, 6], 7, -4, 0, 4, -4, [-16, 20, -24], [-16, 20, -24], 是, -28, -28, 是, x=x_plus-x_minus_with_both_nonnegative_preserves_column_and_objective_contribution)

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。