Principles of Mathematics in Operations Research

Solutions

X

(0.1.0)

ts

Fig. S.8. 3-dimensional pyramid

7.4

See Figure S.8 for a drawing of Pn+\.

Let a^1 be the normal to face Fj, i — 0,1,2,3,4. Let alx < bi be the

respective defining inequalities.

We know Fo is the X\-X2 plane. Then, Fo = {x *E K^3 : x 3 = 0}.

We know that a^2 and a^4 are perpendicular to x 2 --axis. Similarly, a^1 and

aA are perpendicular to xj axis. Thus,

a} = (0,*,*)T, a^2 = (*,0,*)T, a^3 = (0,*,*)^7 ', a^4 = (*,0,*)r.

Since F] contains (1/2,1/2,1), (1,0,0), (0,0,0), what we have is

Fi = {x e R^3 : Oxi - 2x 2 + lz 3 = °} •

Since F 2 contains (1/2,1/2,1), (1,0,0), (1,1,0), we have

F 2 = {x e R^3 : 2xi + 0x 2 + lx 3 = 2}.

Since F 3 contains (1/2,1/2,1), (1,1,0), (0,1,0), it is

F 3 = {x e R^3 : Oxi + 2x 2 + I13 = 2}.

And finally, (1/2,1/2,1), (0,1,0), (0,0,0) are in F 4 ,

F 4 = {x £ R^3 : -2xi + 0x 2 + lx 3 = 0}.

Therefore,

F 3 = {x G R^3 : x 3 > 0, -2x 2 + x 3 < 0, 2xi + x 3 < 2,

2x 2 +x 3 < 2, -2xi +x 3 <0}.