I have an exam in a few hours. I need to understand the solution to the following questionFind the Maximal to the the following 2 x 1 + 3 x 2 is the objective function.The constraints are 4 x 1 + 3 x 2 ≤ 600 ; x 1 + x 2 ≤ 160 ; 3 x 1 + 7 x 2 ≤ 840 ; x 1 , x 2 ≥ 0.Also the answer should be in the "Dual". If its possible please do it in the Algebraic method. If not I would just like the solution using the tableau method and how do you arrive to the solution.(PS: Any help would be great. )

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I have an exam in a few hours. I need to understand the solution to the following questionFind the Maximal to the the following   2      x    1    +  3      x    2   is the objective function.The constraints are  4      x    1    +  3      x    2    ≤  600  ;      x    1    +      x    2    ≤  160  ;  3      x    1    +  7      x    2    ≤  840  ;      x    1    ,      x    2    ≥  0.Also the answer should be in the &quot;Dual&quot;. If its possible please do it in the Algebraic method. If not I would just like the solution using the tableau method and how do you arrive to the solution.(PS: Any help would be great. )

Jenna Farmer · Accepted Answer

First of all, to solve this with the simplex method (tableau method) the inequalities of the contraints should be equalities. So you have to use slack variables.  4      x    1    +  3      x    2    ≤  600      x    1    +      x    2    ≤  160  3      x    1    +  7      x    2    ≤  840      x    1    ,      x    2    ≥  0  ⇓  4      x    1    +  3      x    2    +      x    3    =  600      x    1    +      x    2    +      x    4    =  160  3      x    1    +  7      x    2    +      x    5    =  840      x    1    ,      x    2    ,      x    3    ,      x    4    ,      x    5    ≥  0Then can you continue by creating the tableau?As regards the Algebraic method:Draw the       x    1  -axis and the       x    2  -axis. Then from the first contraint you get:  4      x    1    +  3      x    2    ≤  600  :   when       x    1    =  0  ⇒      x    2    =  200   and  when       x    2    =  0  ⇒      x    1    =  150. So you get the line as below, and since it is   ≤  600 the feasible region is below this lineCan you do this for the other contraints and find the solution? You should also keep in mind that       x    1    ,      x    2    ≥  0

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