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Pertanyaan

Diketahui sistem pertidaksamaan x + 3 y ≤ 9 , 2 x + y ≤ 8 , x ≥ 0 , dan y ≥ 0. Nilai maksimum dari fungsi obyektif f ( x , y ) = 2 x + 3 y adalah ....

  1. 8

  2. 9

  3. 12

  4. 18

  5. 24

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Y. Laksmi

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

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Pembahasan

left parenthesis 1 right parenthesis space space straight x plus 3 straight y equals 9 space  Jika space straight x equals space 0 space space maka space straight y space equals space 3 space space titiknya space didapat space left parenthesis 0 comma 3 right parenthesis space  Jika space straight y equals space 0 space maka space straight x equals space 9 space space titiknya space didapat space left parenthesis 9 comma 0 right parenthesis space    left parenthesis 2 right parenthesis space space 2 straight x plus straight y equals 8 space  Jika space straight x equals space 0 space maka space straight y space equals space 8 space space titiknya space didapat space left parenthesis 0 comma 8 right parenthesis space  Jika space straight y equals space 0 space maka space straight x equals space 4 space space titiknya space didapat space left parenthesis 4 comma 0 right parenthesis space    left parenthesis 3 right parenthesis space space straight x comma straight y space greater or equal than 0 space space Daerah space himpunan space penyelesaiannya space di space kuadran space straight I

Maka space terdapat space 3 space titik space colon  1 right parenthesis thin space Titik space left parenthesis 0 comma 3 right parenthesis  2 right parenthesis thin space Titik space left parenthesis 4 comma 0 right parenthesis  3 right parenthesis thin space Dengan space menggunakan space sistem space eliminasi space space  x plus 3 y equals 9 space space vertical line cross times 2 vertical line space 2 x plus 6 y equals 18 space space space space space  bottom enclose 2 x plus y equals 8 space space vertical line cross times 1 vertical line space 2 x plus space space y equals 8 space space space space minus end enclose  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 5 y equals 10  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y equals 2 space space space space    Substitusi space nilai space straight y space pada space persamaan space left parenthesis 1 right parenthesis space space space space space space space space  x plus 3 left parenthesis 2 right parenthesis equals 9 space space space space space  x equals 3 space space space space space space  Jadi space titik space ketiga space diperoleh space left parenthesis 3 comma 2 right parenthesis space

Cek space setiap space titik space pojok space ke space straight f left parenthesis straight x comma straight y right parenthesis equals 2 straight x plus 3 straight y space  straight f left parenthesis 0 comma 3 right parenthesis equals 0 plus 3 left parenthesis 3 right parenthesis equals 9 space space space  straight f left parenthesis 4 comma 0 right parenthesis equals 2 left parenthesis 4 right parenthesis plus 0 equals 8 space  straight f left parenthesis 3 comma 2 right parenthesis equals 2 left parenthesis 3 right parenthesis plus 3 left parenthesis 2 right parenthesis equals 12 space    Jadi space nilai space maksimumnya space adalah space 12

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Diketahui sistem pertidaksamaan x + 3y ≤ 6, 2x + y ≤ 4, x ≥ 0, y ≥ 0. Nilai maksimum dari fungsi objektif f(x,y) = 4x - 3y adalah ....

18

3.7

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