Lagrange Multiplier problem

Let Lagrange fuction F(x,y,z) = x² + y² + z² + λ(x + 2y + z - 2) + δ(x - y - 7)
then
F'x = x + λ + δ
F'y = y + 2λ - δ
F'z = z + λSolve
x + λ + δ = 0 ... (1)
y + 2λ - δ = 0 ... (2)
z + λ = 0 ... (3)
x + 2y + z - 2 = 0 ... (4)
x - y - 7 = 0 ... (5)(1) + (2) - 3(3) :
x + y - 3z = 0 ... (6)3(4) + (6) :
4x + 7y - 6 = 0 ... (7)7(5) + (7) :
11x - 55 = 0
x = 5
y = - 2
z = 1Let x = ∞+7 , y = ∞ , z = - 3∞ - 5 , then f(x,y,z) is infinity.
So (x , y , z) = (5 , - 2 , 1) is the minimum point of f(x,y,z).
Thus, f(x,y,z) have no maximum and the minimum is f(5 , -2 , 1) = 5² + (- 2)² + 1² = 30.

2014-11-04 12:24:41 補充：
Revision:

F'x = 2x + λ + δ
F'y = 2y + 2λ - δ
F'z = 2z + λ

Solve
2x + λ + δ = 0 ... (1)
2y + 2λ - δ = 0 ... (2)
2z + λ = 0 ... (3)
x + 2y + z - 2 = 0 ... (4)
x - y - 7 = 0 ... (5)

是否：

F'x = 2x + λ + δ
F'y = 2y + 2λ - δ
F'z = 2z + λ

Solve
2x + λ + δ = 0 ... (1)
2y + 2λ - δ = 0 ... (2)
2z + λ = 0 ... (3)

其餘解答正確。
謝謝。