Computer Aided Engineering Design

OPTIMIZATION 357

∂

∂

L

x

x

2

= 2 + = 0 2 λ

and λ[(xx 2 + 2) – 12 ] = 0

Case I: Ifλ = 0, then x 2 = 0 and x 1 = 1 and so g 1 (1, 0) is 1 which is greater than 0. Thus, this solution
is not feasible.

Case II: Forλ≠ 0 we have from the third condition (x 2 + 2) –x 12 = 0 or xx 2 = – 2. 12 Eliminating
λ from the other two conditions gives

(1 + 2x 2 )x 1 – 1 = 0

or 2 xx 13 – 3 – 1 = 0 1

which is a cubic in x 1. The three solutions are 1.36, –1 and –0.36 and the respective values of x 2 are

• 0.13, –1 and –1.86. Since λ = –2x 2 , the corresponding Lagrangian multipliers are 0.27, 2.00 and
  3.73 which are all positive. Thus, (1.36, – 0.13), (–1, –1) and (–0.36, –1.86) are the three local minima
  suggested by the KKT necessary conditions for optimality with function values as 0.15, 5 and 5.35,
  respectively, as depicted by points A, B and C in Figure 12.12. Note that at these points, the function
  and constraint contours are tangent as expected since the function and gradient normals are collinear
  satisfying the relation –∇f = λ∇g for the active constraint. This case is like the minimization of the

objective with an equality (active) constraint. The sufficiency conditions is that H =
∂
∂∂

⎡

⎣

⎢

⎤

⎦

⎥

(^2) L
xxpq
,p=
1, 2,q = 1, 2 should be positive definite. One may compute the Hessian as [2 – 2λ 0; 0 2] and
find that the eigen values are 2(1 – λ) and 2 which are both positive only for A(1.36, –0.13) for λ =
0.27. Thus, (1.36, – 0.13) satisfies both the necessary and sufficient conditions and is a true minimum
while the other two solutions do not satisfy the sufficiency condition.
C
A
B
x 1
x 2
2
1
0
–1
–2
–3 –2 –1 0 1 2 3
Figure 12.12 Function (thin lines) and constraint (thick line) contours for Example 12.9
(b) Consider now an additional constraint g 2 (x 1 ,x 2 )≡ – x 1 ≤ 0 for which the Lagrangian becomes
L = (^1 – 1) + + [( + 2) – ] –
2
2
2
12 1
2
xxxxxλλ 21
and the KKT necessary conditions for an optimum are