Suppose that x, y, z, ν, N, g are some postitive parameters and A, B and C are variables that belong to ( 0 , + ∞ ). I want to identify one or all of the feasible points taht solve the following system of inequalities with respect to A, B and C x A − ν g B − N g C &gt; 0 − ν g A + y B − N g C &gt; 0 − ν g A − ν g B + z C &gt; 0

Question

Suppose that   x,   y,   z,   ν,   N,   g are some postitive parameters and   A,   B and   C are variables that belong to   (  0  ,  +  ∞  ). I want to identify one or all of the feasible points taht solve the following system of inequalities with respect to   A,   B and   C  x  A  −  ν  g  B  −  N  g  C  &amp;gt;  0    −  ν  g  A  +  y  B  −  N  g  C  &amp;gt;  0    −  ν  g  A  −  ν  g  B  +  z  C  &amp;gt;  0

Erzrivalef6 · Accepted Answer

We can rewrite the system of inequalities, including the restrictions on A,B,C to be positive, as follows:                    A                            &amp;gt;                              ν            g                    x                B        +                              N            g                    x                C                            B                            &amp;gt;                              ν            g                    y                A        +                              N            g                    y                C                            C                            &amp;gt;                              ν            g                    z                A        +                              ν            g                    z                B                                          A        ,        B        ,        C                            &amp;gt;        0            The feasible region is convex, an intersection of six open half-spaces, each delineated by a plane passing through the origin. It is possible for this intersection to be empty, unlike the relaxed &quot;weak inequalites&quot; system where the origin (0,0,0) will always be a solution.Note that if (A,B,C) is a feasible point, then so too is any positive scalar multiple r&amp;gt;0 of it, (rA,rB,rC). So geometrically, if the feasible region is not empty, it consists of a collection of open rays proceeding from the origin.If the coefficients were known, we could represent the feasible region by taking its intersection with the plane:  A  +  B  +  C  =  1That intersection (if not empty) is an open polygon (the polygon excluding its perimeter). Such a polygon would be convex and have at most six sides, and its vertices can be readily computed by many linear programming packages. The feasible region of your problem is then all the open rays proceeding from the origin through a point in that open polygon. The points in the open polygon can be parameterized by barycentric coordinates.

Denisse Valdez · Accepted Answer

Solving the system for   A  ,  B  ,  C      {                            A          x          −          B          g          ν          −          C          g          N          =                      e            1            2                                                −          A          g          ν          +          B          y          −          C          g          N          =                      e            2            2                                                −          A          g          ν          −          B          g          ν          +          C          z          =                      e            3            2                                  we get      {                            A          =                                                    e                1                2                                            (                                  g                  2                                N                ν                −                y                z                )                            −              g                              (                                  e                  3                  2                                N                (                g                ν                +                y                )                +                                  e                  2                  2                                ν                (                g                N                +                z                )                )                                                                    g                2                            N              ν              (              2              g              ν              +              x              +              y              )              +              z                              (                                  g                  2                                                  ν                  2                                −                x                y                )                                                                          B          =                                                    e                2                2                                            (                                  g                  2                                N                ν                −                x                z                )                            −                              e                3                2                            g              N              (              g              ν              +              x              )              −                              e                1                2                            g              ν              (              g              N              +              z              )                                                      g                2                            N              ν              (              2              g              ν              +              x              +              y              )              +              z                              (                                  g                  2                                                  ν                  2                                −                x                y                )                                                                          C          =                                                    e                3                2                                            (                                  g                  2                                                  ν                  2                                −                x                y                )                            −                              e                2                2                            g              ν              (              g              ν              +              x              )              −                              e                1                2                            g              ν              (              g              ν              +              y              )                                                      g                2                            N              ν              (              2              g              ν              +              x              +              y              )              +              z                              (                                  g                  2                                                  ν                  2                                −                x                y                )                                                            or assuming       g    2    N  ν  (  2  g  ν  +  x  +  y  )  +  z      (          g      2              ν      2        −    x    y    )    &amp;gt;  0      {                                        A            ′                    =                      e            1            2                                (                          g              2                        N            ν            −            y            z            )                    −          g                      (                          e              3              2                        N            (            g            ν            +            y            )            +                          e              2              2                        ν            (            g            N            +            z            )            )                                                            B            ′                    =                      e            2            2                                (                          g              2                        N            ν            −            x            z            )                    −                      e            3            2                    g          N          (          g          ν          +          x          )          −                      e            1            2                    g          ν          (          g          N          +          z          )                                                  C            ′                    =                      e            3            2                                (                          g              2                                      ν              2                        −            x            y            )                    −                      e            2            2                    g          ν          (          g          ν          +          x          )          −                      e            1            2                    g          ν          (          g          ν          +          y          )                        The matrix associated to the linear system is  M  =      (                            x                          −          g          ν                          −          g          N                                      −          g          ν                          y                          −          g          N                                      −          g          ν                          −          g          ν                          z                      )  and as long as   det      (    M    )    =  x  y  z  −  2      g    3    N      ν    2    −      g    2    ν  (  N  (  x  +  y  )  +  ν  z  )  ≠  0 we have       e    1    ,      e    2    ,      e    3   as independent variations.

Suppose that x , y , z , ν , N , g are some postitiv

Answered question

Answer & Explanation

New Questions in Algebra I