Constraints:
2X1+2X2≤200⇒X1+X2≤100 (C1)
3X3≤150⇒X3≤50 (C2)
4X1+4X3≤600⇒X1+X3≤150 (C3)
X1≥0 (C4)
X2≥0 (C5)
X3<0 (C6)
In 3D the feasible region is the interior of a solid. If the three variables are Cartesian axes, then the solid lies below the X1-X2 plane in the negative region of X3 and in the positive regions of X1 and X2 (C4-C6). C6 supersedes C2, so C2 is redundant. The three planes X1-X2, X1-X3, and X2-X3 form boundaries.
The equations X1+X2=100 (from C1) and X1+X3=150 (from C3) represent boundary lines. On the X1 axis we have two intercepts: one at 100 and the other at 150. The constraints require a point within the boundary plane. The boundary plane from C1 lies within the boundary plane from C3, so the feasibility region is the volume enclosed by the plane from C1. So we can dismiss the boundary plane from C3.
Now we consider Z=50X1+100X2+150X3. C6 implies that X3 must take the smallest possible negative value because we are trying to maximise Z. Since X3<0 we need a value as close to zero as possible. This reduces the problem to 2D (X1 and X2), with X3~0. The 2D region is a right isosceles triangle with vertices at (0,0), (100,0), (0,100). This triangle has infinitesimal thickness (depth).
If we plug in the vertices coordinates into Z we get respectively 0, 5000, 10000, so Z is maximised when X1=0, X2=100, X3=0-. This satisfies all the constraints.