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Risoluzione dei problemi di programmazione lineare - esempio

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  • Note di Apprendimento
  • Revisione degli argomenti
    Amah K.
    NG
    Amah K.

    A+B=9 eqn1. 2A+3B=24 eqn2. Multiply eqn1 by 3 and eqn2 by 1. 3A+3B=27 2A+3B=24 Subtract the result. A=3. Put A=3 in eqn1. 3+B=9 B=9-3=6. Hence,maximum profit=3(35)+6(25)=105+150=255.

    Amah K.
    NG
    Amah K.

    A+B=9 eqn1 2A+3B=24 eqn2. After solving simultanously,A=3,B=6. Therefore,the max. profit=3(35)+6(25)=105+150=255.

    Ахмед Х.
    EG
    Ахмед Х.

    i solve it and i get the max = 315

    Ахмед Х.
    EG
    Ахмед Х.

    my solution http://prntscr.com/8axf7s

    Rathnayaka Mudiyanselage Pasindu L.
    LK
    Rathnayaka Mudiyanselage Pasindu L.

    A + B = 9 ----- (1) 2A + 3B = 24 ----- (2) (1) X 3 = 3A + 3B = 27 ----- (3) (3) - (2) = A = 3 --------------------------------------------------------------------------------------------------------------------------------------------- P = 35A + 25B P = 35 X 3 + 25 X 6 P = 225/=

    Natha R.
    LK
    Natha R.

    maximum profit is 302.3

    Asha N.
    NL
    Asha N.

    the maximum possible profit per week is 255

    Ghanshyam D.
    GB
    Ghanshyam D.

    355 is max profit

    Yasser M.
    EG
    Yasser M.

    255 if it's 315 you will produce no one of type B

    Sophie N.
    GB
    Sophie N.

    315 for A and is none

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