Lagrange Multiplier?

令 g(x,y,z) = x² + 2y² + 3z² - 1
則限定條件為 g = 0

由 Lagrange 乘數法得 :
∇ f = λ * ∇ g
2x i - 2y j + 2z k = λ ( 2x i + 4y j + 6z k )
2x = 2xλ , - 2y = 4yλ , 2z = 6zλ
2x( 1 - λ ) = 0 , 2y( 1 + 2λ ) = 0 , 2z( 1 - 3λ ) = 0

(1) 當 λ = 1
y = z = 0
g = x² - 1 = 0
x² = 1
f = 1

(2) 當 λ = - 1/2
x = z = 0
g = 2y² - 1 = 0
y² = 1/2
f = - 1/2

(3) 當 λ = 1/3
x = y = 0
g = 3z² - 1 = 0
z² = 1/3
f = 1/3

(4) 當 λ ≠ 1 , - 1/2 , 1/3
x = y = z = 0
g = - 1 = 0 , 矛盾

max = max { 1 , - 1/2 , 1/3 } = 1
min = min { 1 , - 1/2 , 1/3 } = - 1/2

Ans: max = 1 , min = - 1/2