The following problem max x , y f ( x , y ) = log ⁡ x + log ⁡ ( x + y ) − 2 x − 3 ysubject to x ≥ 0 , y ≥ 0is a concave maximization problem, and thus can be numerically solved by convex optimization methods. The optimal solution may be ( x ∗ , y ∗ ) = ( 1 , 0 ).However, I want to derive the solution analytically. I tried the following approach, for example, by taking partial derivative, ∂ f ∂ x = 1 x + 1 x + y − 2 ∂ f ∂ x = 1 x + y − 3And there exists no solution that makes both partial derivative to be 0.

Question

The following problem      max          x      ,      y        f  (  x  ,  y  )  =  log  ⁡  x  +  log  ⁡  (  x  +  y  )  −  2  x  −  3  ysubject to  x  ≥  0  ,  y  ≥  0is a concave maximization problem, and thus can be numerically solved by convex optimization methods. The optimal solution may be   (      x    ∗    ,      y    ∗    )  =  (  1  ,  0  ).However, I want to derive the solution analytically. I tried the following approach, for example, by taking partial derivative,            ∂      f              ∂      x        =      1    x    +      1          x      +      y        −  2            ∂      f              ∂      x        =      1          x      +      y        −  3And there exists no solution that makes both partial derivative to be 0.

feaguelaBapzo · Accepted Answer

What you need for optimisation under constraints are the Karush-Kuhn-Tucker conditions, which, for your problem can be written as            ∂              L                    ∂      x            (    x    ,    y    ,    μ    )    =  0              ∂              L                    ∂      y            (    x    ,    y    ,    μ    )    =  0    x  &amp;gt;  0  ,  y  ≥  0  ,  μ  ≥  0    y  μ  =  0     ,where          L      is given by      L        (    x    ,    y    ,    μ    )    =  f  (  x  ,  y  )  +  μ  y     .With          L      thus defined, we have                                          ∂                          L                                            ∂            x                                    (          x          ,          y          ,          μ          )                                    =                  1          x                +                  1                      x            +            y                          −        2        =        0                                                  ∂                          L                                            ∂            x                                    (          x          ,          y          ,          μ          )                                    =                  1                      x            +            y                          −        3        +        μ        =        0                            .                     From this, bit of elementary algebra gives  x  =      1          μ      −      1           ,which implies      μ  &amp;gt;  1   , since      x  &amp;gt;  0   , and hence      y  =  0   , from the condition      μ  y  =  0   . Substituting      y  =  0    in the first of the above conditions then gives       x  =  1   , and thence,      μ  =  2   .

The following problem <munder> <mo movablelimits="true" form="prefix">max <mrow cla

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