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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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    You probably meant $x,y,z \ge 0$. – 2017-02-02
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    @quasi yes, updated it now – 2017-02-02
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    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. – 2017-02-02
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    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. – 2017-02-02
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    @Moo: Yes. I finally decided "enough is enough" with all the cranks and trolls that infest sci.math. But I still lurk there. – 2017-02-02
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    Also, with the graph of the plane F = 72 against the feasible region, it would be nice to also show an outward normal vector to the plane F = 72, at the point where the plane F = 72 touches the feasible region. – 2017-02-02
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    @quasi Is there a free alternative software that I can use to graph and how? I need an Idea so how the region looks so I can graph It by hand. – 2017-02-02
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    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 $(6,0,0)$. – 2017-02-02
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    To graph those planes, just find the intercepts for each, and draw the triangle with the intecepts as vertices. – 2017-02-02
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    @quasi Thank you so much for your help ! – 2017-02-02

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