Computer Aided Engineering Design

352 COMPUTER AIDED ENGINEERING DESIGN

LL L( + , + ) – ( , ) =^1

2

00 00 T (, ) + ()

2

2 00

T

XX X X 0

X

XX

X

ΔΔΛΛΛΛΛΛΛΔ∂ ΛΛΔΔΛ GX ΔX

∂

∂

∂

⎧

⎨

⎩

⎫

⎬

⎭

Expanding the constraints about the extremum and using the necessary condition for optimality

GX X GX

X

GX X

X

( 000 0 + ΔΔΔ) = ( ) + ∂ ( ) = GX( ) X

∂

∂

∂ (12.18)

ForΔX satisfying all the constraints, G(X 0 + ΔX) = 0 and so ∂
∂X
GX() =. 0 ΔX 0 Thus

LL L( + , + ) – ( , ) =^1

2

000 T (, )

2

XX 0 ΔΔΛΛΛΛΛXΛΛΔXX 2 XX 00 Λ Δ

∂

∂

(12.19)

and for the left hand side to be greater than 0 at a local minimum,
∂
∂

2

X^2 L(, )X^00 ΛΛ

must be positive

definite.

Example 12.7. Calculate the maximum and minimum values of f = x^2 + y^2 subject to g≡x^2 + y^2 + 2x

• 2y + 1 = 0 using the Lagrangian multiplier method.
  We form the Lagrangian as
  L = x^2 + y^2 + λ(x^2 + y^2 + 2x – 2y + 1)

differentiating which with respect to the variables x and y, respectively, gives

2 + (2 + 2) = 0 = –

+ 1

xxλ xλ

λ

⇒

2 + (2 – 2) = 0 =

+ 1

yyλ yλ

λ

⇒

Substituting in the constraint and rearranging gives

λ^2 + 2λ – 1 = 0 ⇒ λ = –1 ±√ 2

Thus, for λ = –1 + √2,x= 1/√2 – 1 and y = –1/√2 + 1, and for λ = –1 – √2,x = – (1/√2 + 1) and
y = (1/√2 + 1). The Hessian matrix is

H = =

2 + 2 0

02 + 2

2

2

2

(^22)
2
∂
∂
∂
∂∂
∂
∂∂
∂
∂
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎤
⎦
⎥
LL
L L
x xy
xy y
λ
λ
with eigenvalues as 2λ + 2. For λ = – 1 + √2, the eigenvalues are 2√2 which are positive and hence
the Hessian is positive definite. For λ = –1 – √2, however, the eignvalues are – 2√2 and negative, the
Hessian for which is negative definite and the point is a relative maximum. Figure 12.9 shows the
graphical depiction of the optimal solutions for this example. The function contours are shown in thin
lines which are tangent to the constraint curve (thick circle) at points A and B, where A yields the
function minimum and B gives the maximum. The corresponding function values at A and B are
3 – 2 2 and 3 + 2 2.