I have to maximize the following function -Max A C 1 − m / − m + (1-A) C 2 − m / − m Subject to, C 1 ≤ 5(1-x) + x C 2 ≤ 3(1-x) + 7x1≤x≤10I wrote it as: L(x) = f(x) - λ 1 ( C 1 - 5(1-x) + x) - λ 2 ( C 2 - 3(1-x) + 7x) - λ 3 (x-1) - λ 4 (x-10)Can I write last constraint as partioned into λ 3 and λ 4 . Is there some other way to introduce such box constraints into the same problem?

Question

I have to maximize the following function -Max A       C    1          −      m            /        −    m   + (1-A)       C    2          −      m            /        −    m  Subject to,      C    1   ≤ 5(1-x) + x      C    2   ≤ 3(1-x) + 7x1≤x≤10I wrote it as: L(x) = f(x) -             λ        1  (      C    1   - 5(1-x) + x) -             λ        2  (      C    2   - 3(1-x) + 7x) -             λ        3  (x-1) -             λ        4  (x-10)Can I write last constraint as partioned into             λ        3   and             λ        4  . Is there some other way to introduce such box constraints into the same problem?

Elliana Shelton · Accepted Answer

Considering  L  (  x  ,  λ  ,  ϵ  )  =      λ    1    (      C    1    −  5  (  1  −  x  )  −  x  −      ϵ    1    2    )  +      λ    2    (      C    2    −  3  (  1  −  x  )  −  7  x  −      ϵ    2    2    )  +      λ    3    (  1  −  x  −      ϵ    3    2    )  +      λ    4    (  x  −  10  −      ϵ    4    2    )with       ϵ    k   convenient slack variables, the stationary points are the solutions for  ∇  L  =  0  =      {                            4                      λ            1                    −          4                      λ            2                    −                      λ            3                    +                      λ            4                    =          0                                                  C            1                    −          5          (          1          −          x          )          −          x          −                      ϵ            1            2                    =          0                                                  C            2                    −          3          (          1          −          x          )          −          7          x          −                      ϵ            2            2                    =          0                                      1          −          x          −                      ϵ            3            2                    =          0                                      x          −          10          −                      ϵ            4            2                    =          0                                                  λ            1                                ϵ            1                    =          0                                                  λ            2                                ϵ            2                    =          0                                                  λ            3                                ϵ            3                    =          0                                                  λ            4                                ϵ            4                    =          0                        and the feasible solutions are those observing       ϵ    k    2    ≥  0 for   k  =  1  ,  2  ,  3  ,  4      (                            x                                      λ            1                                                λ            2                                                λ            3                                                λ            4                                                ϵ            1            2                                                ϵ            2            2                                                ϵ            3            2                                                ϵ            4            2                                                                          5              −                              C                1                                      4                                    0                          0                          0                          0                          0                                      C            1                    +                      C            2                    −          8                                                                    C                1                            −              1                        4                                    −                      1            4                    (                      C            2                    +          35          )                                                                                C                2                            −              3                        4                                    0                          0                          0                          0                                      C            1                    +                      C            2                    −          8                          0                                                    7              −                              C                2                                      4                                                                              C                2                            −              43                        4                                                x                          0                          0                          0                          0                                      C            1                    +          4          x          −          5                                      C            2                    −          4          x          −          3                          1          −          x                          x          −          10                      )

agrejas0hxpx · Accepted Answer

Is your function constant? If I read correctly you have                              maximise                      w.r.t                        x            ∈                          R                                                                          −        A                                                    C              1                              −                m                                      m                          −        (        1        −        A        )                                            C              2                              −                m                                      m                          =        γ                            subj.to                                      h          1                (        x        )        ≤        0                                                    h          2                (        x        )        (        x        )        ≤        0                                          α        ≤                  h          3                (        x        )        ≤        β            Maximizing a constant function under some constraints is called a satisfiability problem; any reachable value will attain the maximum (and minimum) of   γ. Your method of putting  α  ≤      h    3    (  x  )  ≤  β      ⟺      α  −      h    3    (  x  )  ≤  0    and        h    3    (  x  )  −  β  ≤  0Before putting the problem into Lagrangian form is the go-to solution in most cases if   f is non constant!

I have to maximize the following function - Max A <msubsup> C 1 <mrow class="M

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