Principles of Mathematics in Operations Research

232 Solutions

Then, we have

V^2 /(*A)

32

2 1

and V^2 f(xB) =

52

2 1

Let us check the positive definiteness of V^2 f(xA) using the definition:

vTV^2 f{xA)v = [vi,vi]

32

2 1 Zv{ + 4«iv^2 + v\

If Vl = -0.5 and v 2 = 1.0, we will have vTV^2 /(xA)w < 0. On the other hand,
if vi = 1.5 and v 2 = 1.0, we will have vTV^2 f(xA)v > 0. Thus, S7^2 f(xA) is
indefinite. Let us check V^2 /(XB):

vTV^2 /(xB)w = [vi,i> 2 ]
5 2
2 1

= 5v^2 + 4viv 2 + v\ = v\ + (2wi + v 2 )^2 > 0.

Thus, V^2 /(XB) is positive definite and xB

f{xB) = 19.166667.

is a local minimizer with

Fig. S.5. Plot of f(xi,x 2 ) = \x\ + \x\ + 2xix 2 + \x\ - x 2 + 19