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L ′ y (x, y, λ) = ex+y − 3 − 2x + λ
L ′ λ (x, y, λ) = x + y − 2 (TOUJOURS L′ λ (x, y, λ)= g(x, y))
2 ème étape : les points critiques
L ′ x (x, y, λ) = 0 ex+y − 3 + 2x − 2y + λ = 0 1
L ′ y (x, y, λ) = 0 ⇨ ex+y − 3 − 2x + λ = 0 2
L ′ λ (x, y, λ) = 0 x + y − 2 = 0 3
1 - 2 = 4x − 4y ⟺ 4x = 4y ⟺ x = y
3 ⟹ 2x − 2 = 0 ⟺ 2x = 2 ⟺ x = 1 et y = 1
1 ⟹ e 2 − 3 + 2 − 2 + λ = 0 ⟺ λ = 3 − e 2
⟹ A(1, 1, 3 − e 2 )
3 ème étape : les dérivées partielles secondes
On note
L ′′ x 2(x, y, λ) L′′ yx (x, y, λ) g′ x (x, y)
H =
L ′′ y2(x, y, λ) g′
y (x, y)
0
g′ x (x, y) = 1
g′ y (x, y) = 1