Let f ( x 1 , … , x n ) = x 1 x 2 ⋅ ⋅ ⋅ x n . Let A := { x ∈ R n : x 1 + x 2 + ⋯ + x n = n , x i ≥ 0 ∀ i }. I want to find the global max of f under the constrain given by A. MY approach:Put F ( x 1 , … , x n ) = ∑ i = 1 n x i − n = 0 is the constrain. We use lagrange multipliers: ∇ f = λ ∇ F ⟹ ( x 2 ⋅ ⋅ ⋅ x n , … , x 1 ⋯ x n − 1 ) = ( λ , λ , … , λ ) .Hence we have x 2 ⋅ ⋅ ⋅ x n = x 1 x 3 ⋅ ⋅ ⋅ x n = ⋯ = x 1 ⋅ ⋅ ⋅ x n − 1 this implies that x 1 = x 2 = ⋯ = x n We also know ∑ x i = n ⟹ n X = n ⟹ X = 1 if we let X = x i So, max occurs at ( 1 , 1 , … , 1 ) which is f ( 1 , 1 , … , 1 ) = 1 Is this correct? I feel im missing something. thanks for any feedback.

Question

Let   f  (      x    1    ,  …  ,      x    n    )  =      x    1        x    2    ⋅  ⋅  ⋅      x    n  . Let   A  :=  {  x  ∈            R        n    :      x    1    +      x    2    +  ⋯  +      x    n    =  n  ,          x    i    ≥  0      ∀  i  }. I want to find the global max of   f under the constrain given by   A. MY approach:Put   F  (      x    1    ,  …  ,      x    n    )  =      ∑          i      =      1        n        x    i    −  n  =  0 is the constrain. We use lagrange multipliers:  ∇  f  =  λ  ∇  F    ⟹    (      x    2    ⋅  ⋅  ⋅      x    n    ,  …  ,      x    1    ⋯      x          n      −      1        )  =  (  λ  ,  λ  ,  …  ,  λ  )  .Hence we have      x    2    ⋅  ⋅  ⋅      x    n    =      x    1        x    3    ⋅  ⋅  ⋅      x    n    =  ⋯  =      x    1    ⋅  ⋅  ⋅      x          n      −      1      this implies that      x    1    =      x    2    =  ⋯  =      x    n  We also know   ∑      x    i    =  n    ⟹    n  X  =  n    ⟹    X  =  1 if we let   X  =      x    i  So, max occurs at   (  1  ,  1  ,  …  ,  1  ) which is   f  (  1  ,  1  ,  …  ,  1  )  =  1 Is this correct? I feel im missing something. thanks for any feedback.

humbast2 · Accepted Answer

For a rigorous argument you should observe that you also have a boundary where clearly there cannot be maximum(because the function is 0 in that case). So since there is a maximum(because your set is compact), and is not in the boundary, it has to be an inner one. So you can go on with your argument.

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