一個有關 Lagrange 乘數法 的問題

Lagrange's Multiplier:
設F(x,y, t)=(x^4+y^4-4xy+1)+ t(x^2+y^2-k), 0 <= k <= 2
∂F/∂x =0= 4x^3 -4y + 2tx ----(A)
∂F/∂y =0= 4y^3 -4x + 2ty ----(B)
∂F/∂t =0= x^2+y^2-k
y*(A)-x*(B): 4xy(x^2-y^2)+ 4(x^2-y^2) =0
4(x^2-y^2)(xy+1)=0
x^2=y^2 or xy= -1
又x^2+y^2=k, 故 (x^2, y^2)= (k/2, k/2) or (x,y)=(1, -1), (-1, 1)
case1: (x^2, y^2)=(k/2, k/2) 代入f(x,y)=k^2 /2 +/- 2k+1= (1/2) (k +/- 2)^2 -1
最大值為 (1/2)*(4)^2-1= 7 , 最小值= -1
case2: (x,y)= +/- (1, -1)時, 代入f(x,y)得7
故f(x,y)最大值為 7 (此時 (x,y)=+/-(1, -1)), 最小值為 -1 (此時 (x,y)=+/- (1,1) )

另法:改為極坐標
x^2+y^2 <= 2, 設 (x,y)=(rcost, rsint), 0<= r <= √2, t=0~2π
代入 f(x,y)得 r^4[(cost)^4+(sint)^4] - 4r^2 cost sint +1
=r^4[ 1-2(sint cost)^2] - 2r^2 sin(2t) +1
=r^4[ 1-(1/2)(sin 2t)^2] - 2r^2 sin(2t)+1
(配方)= -(1/2)[ r^2 sin(2t) + 2]^2 + r^4 + 3
故 r^2 sin(2t)= -2, r=√2 時, 有最大值為 0+ (√2)^4+ 3= 7
r^2 sin(2t)=r^2 時, f(x,y)= -(1/2)(r^2+2)^2 +r^4+3 有最小值為 -1(此時 r^2= 2)

用slack variable 啊