
Sec. 5.3 The Geometry of Quadratic Functions 91
To provide algebraic verification of the optimality conditions for
Example 5.2, we observe that
g(x) =
−10 + 2 x
1
−6 + 2x
2
so that g(x
0
) = (−2, −2)
′
. The cone part of the optimality conditions
requires that there is a nonnegative number u
2
with
−g(x
0
) = u
2
a
2
.
These are precisely the 2 linear equations in 1 unknown
u
2
= 2,
u
2
= 2,
which have the obvious solution u
2
= 2. Since u
2
is nonnegative, we
have algebraically verified that the optimality conditions are satisfied.
Finally, the optimal solution x
0
= (3, 2)
′
for Example 5.3 lies in the
interior of R. Since no constraint s are active at x
0
, the cone part of the
optimalit ...