82 Classical Optimization Techniques
In terms of the notation of our equations, let us take the independent variables asx 3 =y 3 and x 4 =y 4 so that x 1 =y 1 and x 2 =y 2Then the Jacobian of Eq. (2.27) becomesJ
(
g 1 , g 2
x 1 , x 2)
=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂g 1
∂y 1∂g 1
∂y 2
∂g 2
∂y 1∂g 2
∂y 2∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
=
∣
∣
∣
∣
12
1 2
∣
∣
∣
∣=^0
and hence the necessary conditions of Eqs. (2.26) cannot be applied.
Next, let us take the independent variables asx 3 =y 2 andx 4 =y 4 so thatx 1 =y 1
andx 2 =y 3. Then the Jacobian of Eq. (2.27) becomesJ
(
g 1 , g 2
x 1 , x 2)
=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂g 1
∂y 1∂g 1
∂y 3
∂g 2
∂y 1∂g 2
∂y 3∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
=
∣
∣
∣
∣
1 3
1 5
∣
∣
∣
∣=^2 =^0
and hence the necessary conditions of Eqs. (2.26) can be applied. Equations (2.26) give
fork=m+ 1 = 3
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂f
∂x 3∂f
∂x 1∂f
∂x 2
∂g 1
∂x 3∂g 1
∂x 1∂g 1
∂x 2
∂g 2
∂x 3∂g 2
∂x 1∂g 2
∂x 2∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂f
∂y 2∂f
∂y 1∂f
∂y 3
∂g 1
∂y 2∂g 1
∂y 1∂g 1
∂y 3
∂g 2
∂y 2∂g 2
∂y 1∂g 2
∂y 3∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ =
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
y 2 y 1 y 3
2 1 3
2 1 5∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
=y 2 ( 5 − 3 )−y 1 ( 01 − 6 )+y 3 ( 2 − 2 )
= 2 y 2 − 4 y 1 = 0 (E 4 )and fork=m+ 2 =n=4,
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂f
∂x 4∂f
∂x 1∂f
∂x 2
∂g 1
∂x 4∂g 1
∂x 1∂g 1
∂x 2
∂g 2
∂x 4∂g 2
∂x 1∂g 2
∂x 2∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂f
∂y 4∂f
∂y 1∂f
∂y 3
∂g 1
∂y 4∂g 1
∂y 1∂g 1
∂y 3
∂g 2
∂y 4∂g 2
∂y 1∂g 2
∂y 3