Math Problem Statement
max(Z)=6⋅x+10⋅y x≤8 y≤8 2⋅x+2⋅y≤20 1⋅x+3⋅y≤15 x,y≥0 Nun wird die Nebenbedingung 1⋅x+3⋅y≤15 durch eine andere ersetzt: Neu gilt 1⋅x+3⋅y≥15 Was ist der optimale Zielfunktionwert? Runden Sie auf ganze Zahlen (keine Nachkommastellen).
Solution
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Math Problem Analysis
Mathematical Concepts
Linear Programming
Constraint Optimization
Inequalities
Formulas
Linear objective function
Theorems
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Suitable Grade Level
Grades 11-12
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