I'm trying to solve this question:
Сonsider the following Polyhedron which has five edges $BCD, BCEO, BDFO, CDFE, EFO$. Write down a canonical set of constraints that define the Polyhedron.
The solution is:
The canonical system of constraints that define the Polyhedron is the triangle $AEF$ that defines the constraint $4x+2y+z\leq 4$ and the triangle $BHG$ that defines the constraint $x+2y+2z\leq 6$ as well as the non-negativity constraints of $x ,y,z$: $$ \begin{align*} &4x+2y+z\leq4\\&x+2y+2z\leq6\\&x,y,z\geq0 \end{align*} $$
I don't understand the answer. I have those questions:
- Why $AEF$ and $BHG$ define the constraints of the Polyhedron? It has five edges like mentioned above so I would think that I would need to find the space of each edge.
- How did they figure technically the constraints? Why the $AEF$ triangle defines $4x+2y+z\leq4$. I know that $A=(0,0,4)$, $E=(1,0,0)$ and $F=(0,2,0)$ so I would assume it should be $x+2y+4z$ and not $4x+2y+z$. Also where the $\leq 4$ came from?
I see the similar question here, but the I still don't understand the answer. The following said there:
Every plane, with equation $ax+by+cz+d=0$, divides all space into two parts (half-spaces) defined by $ax+by+cz+d>0$ and $ax+by+cz+d<0$. Your polyhedron is the intersection of five half-spaces, originating from its five faces.
But technically, how can I get the expected answers?
You need to find equations for the five planes that contain the polyhedron's five faces. Three of the planes are obvious: $x=0$ for BDFO, $y=0$ for BCEO, and $z=0$ for EFO.
Now let's find the equation for the plane containing CDEF. Three points uniquely define a plane through them, so we need the coordinates of three points in that plane. We have coordinates for E and F. We could calculate the exact coordinates of for example C and then find the equation that is satisfied by the coordinates of E, F, and C. However, this is a lot of work.
We know the plane contains the line EC because it contains the points E and C. That line also goes through the point A, so therefore the plane we are looking for also contains point A. We know the coordinates of A, so the plane that contains the face CDEF contains the three points E, F, and A which all have known coordinates.
We have $$E: (1,0,0)\\F: (0,2,0)\\A: (0,0,4)$$
You won't always have points with so many zero coordinates, so you would then need to use the general equation $ax+by+cz=d$, for each point substitute its $x,y,z$ values to get an equation with the unknowns $a,b,c,d$, and solve those three equations. Here we get the three equations $a=d$ from point E, $2b=d$ from F, and $4c=d$ from A. We have four unknowns and only three equations, so we have one degree of freedom left. We can choose a value for $d$, and then this determines $a,b,c$.
Choosing $d=1$ we find that the equation through those points is $x+\frac{y}2+\frac{z}4=1$. We can scale this by a factor of $4$ to remove the fractions to get $4x+2y+z=4$, or we could have chosen $d=4$ to get there directly.
You can do the same for the face BCD. We don't know the coordinates of C or D, and although we could calculate them that is too much work. The plane that contains this face contains the line through BD and the line through BC so contains the points G and H as well as B, all of which have known coordinates. The equation through these three points is $\frac{x}6+\frac{y}3+\frac{z}3=1$ or $x+2y+2z=6$.
These five equations define the planes bounding the polyhedron. To define the space inside the polyhedron you need to change these equations into inequalities, because as explained in the part you quoted, each plane splits space in two, and the coordinates on one side of the plane satisfy the $<$ inequality and the other side satisfies the $>$ inequality.
To determine which direction each of the five inequalities should go, you can take any point that definitely lies inside the polyhedron, for example $(0.1,0.1,0.1)$. We know the coordinates of that point must satisfy all the inequalities, so substitute it into each equation and see whether the left or the right hand side is largest, and change the $=$ into $>$ or $<$ to make it true. These strict inequalities define only the interior of the polygon, so if you want to include the boundary, the faces themselves, you should use $\ge$ and $\le$ instead.