{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-convex-optimization-problem"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-convex-optimization-problem-01","title":{"zh-Hant":"凸最佳化問題：資格核對、可行性、單點求值與兩候選比較，以及它們不能推出什麼","en":"Convex optimization problem: qualification, feasibility, single-point evaluation and a two-candidate comparison, and what they do not imply"},"summary":{"zh-Hant":"問題 P：在可行域 X=[0,5] 上最小化 f(x)=x²−6x+10。（一）分別核對可行域凸與目標凸，判定 P 具凸問題資格；（二）判定候選 x₁=1、x₂=4、x₃=−1 的可行性；（三）求 f(1)、f(4) 並比較兩個可行候選；（四）說明兩候選中較低者不因此是 P 的最優解，並以 P′：在 X′=(0,1] 上最小化 x² 說明凸問題不一定有最小值點。","en":"Problem P: minimize f(x)=x²−6x+10 over the feasible domain X=[0,5]. (1) Check separately that the domain and the objective are convex, so that P qualifies as a convex problem; (2) decide feasibility of the candidates x₁=1, x₂=4, x₃=−1; (3) evaluate f(1) and f(4) and compare the two feasible candidates; (4) explain why the candidate with the lower objective value is not thereby an optimal solution of P, and use P′: minimize x² over X′=(0,1], to show that a convex problem need not have a minimizer."},"mko_ids":["mko-convex-optimization-problem"]}],"title":"Concept derivation guides"}