JEE Main & Advanced Mathematics Linear Programming Question Bank Critical Thinking

  • question_answer Minimize \[z=\sum\limits_{j=1}^{n}{{}}\sum\limits_{i=1}^{m}{{{c}_{ij}}\,{{x}_{ij}}}\]                 Subject to : \[\sum\limits_{j=1}^{n}{{{x}_{ij}}\le {{a}_{i}},\ i=1,.......,m}\]                                   \[\sum\limits_{i=1}^{m}{{{x}_{ij}}={{b}_{j}},\ j=1,......,n}\]                 is a (L.P.P.) with number of constraints                    [MP PET 1999]

    A)                 \[m+n\]              

    B)                 \[m-n\]

    C)                 mn        

    D)                 \[\frac{m}{n}\]

    Correct Answer: A

    Solution :

               Condition (i),                    \[i=1,{{x}_{11}}+{{x}_{12}}+{{x}_{13}}+.....+{{x}_{1n}}\]                                 \[i=2,{{x}_{21}}+{{x}_{22}}+{{x}_{23}}+......+{{x}_{2n}}\]                    \[i=3,{{x}_{31}}+{{x}_{32}}+{{x}_{33}}+......+{{x}_{3n}}\]                    ....................                    \[i=m,{{x}_{m1}}+{{x}_{m2}}+{{x}_{m3}}+.....{{x}_{mn}}\to \]constraints                    Condition (ii),                    \[j=1,\,{{x}_{11}}+{{x}_{21}}+{{x}_{31}}+......+{{x}_{m1}}\]                    \[j=2,{{x}_{12}}+{{x}_{22}}+{{x}_{32}}+......+{{x}_{m1}}\]                    ....................                    \[j=n,{{x}_{1n}}+{{x}_{2n}}+{{x}_{3n}}+......+{{x}_{mn}}\to n\] constraints                                        \ Total constraints = \[m+n\].

adversite


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