Are the following range of values on x 3 correct given a system of inequalities with 3 variables?Given the system of inequalities: x 3 ≤ x 1 + x 2 − 2 2 x 3 ≥ x 1 + 3 x 2 x 3 ≥ 1 − x 2 − 11 3 ≤ x 2 ≤ − 3 13 ≤ x 1 &lt; ∞ I know that x 1 ∈ [ l o w e r x 1 , u p p e r x 1 ) = [ 13 , ∞ ) and x 2 ∈ [ l o w e r x 2 , u p p e r x 2 ] = [ − 11 / 3 , − 3 ].In order to find l o w e r x 3 , u p p e r x 3 , I did the following: l o w e r x 3 = m a x { x 1 + 3 x 2 , 1 − x 2 } = { u p p e r x 1 + 3 u p p e r x 2 , 1 − u p p e r x 2 } = { ∞ , 4 } = ∞ u p p e r x 3 = l o w e r x 1 + l o w e r x 2 − 2 2 = 13 − 11 / 3 − 2 2 = 11 3 Is the above calculation correct? In general, is u p p e r x n = f ( l o w e r x 1 , l o w e r x 2 , … , l o w e r x n − 1 ) and is l o w e r x n = g ( u p p e r x 1 , u p p e r x 2 , … , u p p e r x n − 1 ), where f and g are some real-valued functions?

Question

Are the following range of values on       x    3   correct given a system of inequalities with 3 variables?Given the system of inequalities:                              x          3                                    ≤                                            x              1                        +                          x              2                        −            2                    2                                              x          3                                    ≥                  x          1                +        3                  x          2                                              x          3                                    ≥        1        −                  x          2                                    −                  11          3                                    ≤                  x          2                ≤        −        3                            13                            ≤                  x          1                &amp;lt;        ∞            I know that       x    1    ∈  [  l  o  w  e      r            x      1        ,  u  p  p  e      r            x      1        )  =  [  13  ,  ∞  ) and       x    2    ∈  [  l  o  w  e      r            x      2        ,  u  p  p  e      r            x      2        ]  =  [  −  11  /  3  ,  −  3  ].In order to find   l  o  w  e      r            x      3        ,  u  p  p  e      r            x      3      , I did the following:  l  o  w  e      r            x      3        =  m  a  x      {          x      1        +    3          x      2        ,    1    −          x      2        }    =      {    u    p    p    e          r                  x        1              +    3    u    p    p    e          r                  x        2              ,    1    −    u    p    p    e          r                  x        2              }    =      {    ∞    ,    4    }    =  ∞  u  p  p  e      r            x      3        =            l      o      w      e              r                        x          1                    +      l      o      w      e              r                        x          2                    −      2        2    =            13      −      11          /      3      −      2        2    =      11    3  Is the above calculation correct? In general, is   u  p  p  e      r            x      n        =  f  (  l  o  w  e      r            x      1        ,  l  o  w  e      r            x      2        ,  …  ,  l  o  w  e      r            x              n        −        1              ) and is   l  o  w  e      r            x      n        =  g  (  u  p  p  e      r            x      1        ,  u  p  p  e      r            x      2        ,  …  ,  u  p  p  e      r            x              n        −        1              ), where   f and   g are some real-valued functions?

Haylie Cherry · Accepted Answer

What you&#039;ve written so far is that       x    3    ≥  ∞ and       x    3    ≤      11    3   which doesn&#039;t make sense. In order to figure out the lower and upper bounds for       x    3  , ask yourself, &quot;What is the largest value that       x    3   can be?&quot; Since       x    3    ≤                    x        1            +              x        2            −      2        2  , we know that      x    3    ≤            max      (              x        1            )      +      max      (              x        2            )      −      2        2    =  ∞So the upper bound for       x    3   is   ∞.To find the lower bound, we ask, &quot;What is the smallest value that       x    3   can be?&quot; Since       x    3    ≥      x    1    +  3      x    2   and       x    3    ≥  1  −      x    2  , we can conclude the following two statements:      x    3    ≥  min  (      x    1    )  +  3  min  (      x    2    )  =  13  +  3      (    −          11      3        )    =  2      x    3    ≥  1  −  max  (      x    2    )  =  1  −      (    −    3    )    =  4Notice we use a max in the second inequality because it will produce the smallest value.Therefore it is true that       x    3    ≥  4 and       x    3    ≥  2. Since       x    3    ≥  4 is more restrictive, we use that as our lower bound. So the lower bound for       x    3   is   4.

Are the following range of values on x 3 </msub> correct given a system of inequaliti

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