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Here is the problem:

Maximize: $ F= 12x+12y+5z$

Subject to the constraints:

$2x+4y+3z\leq12$

$x+2y+z\leq8$

$ x,y,z \geq 0$

Using simplex method I got these answers:

$x,y,z(6,0,0), F=72; $

I need to use the graphical method to solve this, but I have no idea how if it contains 3 variables.

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  • $\begingroup$ You probably meant $x,y,z \ge 0$. $\endgroup$ Commented Feb 2, 2017 at 16:52
  • $\begingroup$ @quasi yes, updated it now $\endgroup$ Commented Feb 2, 2017 at 16:54
  • $\begingroup$ The contraints give bounding planes. Graph each plane. It can be done by hand, but I would use a software package such as Maple, Mathematica, Matlab. So now you have a view of the feasible region. Then verify that the plane F = 72 bounds the feasible region and just touches it. $\endgroup$ Commented Feb 2, 2017 at 17:00
  • $\begingroup$ The 3D graphical solution is typically a tedious task, and the final visualization gives visual confirmation of the optimal value, but on its own, without algebraic verification, it's not entirely convincing. If you're forced to do it once or twice, no big deal -- the work involved will make you even more appreciative of the simplex method. $\endgroup$ Commented Feb 2, 2017 at 17:07
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    $\begingroup$ To graph it by hand, just graph the planes \begin{align*} 2x+4y+3z&=12\\ x+2y+z&=8\\ 12x+12y+5z&=72 \end{align*} And to show that $F=72$ is optimal, draw a normal vector (a vector proportional to the vector ${<}12,12,5{>}$), placed at the point $(6,0,0)$. $\endgroup$ Commented Feb 2, 2017 at 17:27

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Normally if one were to draw this by hand, one would change all the inequalities into equalities, then graph those planes, mark the direction where the inequalities imply, denote a feasible region, and then take note of where the intersections happen between the constraints in the feasible region.

Graphically, after doing all of this, the feasible region of the model would look like the following:

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All the extreme points (shown in red in the following picture), are

  • $\left(0,0,0\right) \longrightarrow z = 0$
  • $\left(6,0,0\right) \longrightarrow z = 72$
  • $\left(0,3,0\right) \longrightarrow z = 36$
  • $\left(0,0,4\right) \longrightarrow z = 20$

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Since we're looking to maximize $z$, the optimal solution for the model would be $(6,0,0)$ where $z=72$.

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