作者alasa15 (alasa)
看板Math
标题[微积] Lagrange 乘数法 不等式型限制
时间Wed Jul 6 22:59:06 2011
Find the extreme value of f(x,y)= x^4+y^4-4xy+1
subject to the constraint x^2+y^2<=2
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Lagrange's Multiplier:
设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
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y*(A)-x*(B): 4xy(x^2-y^2)+ 4(x^2-y^2) =0
4(x^2-y^2)(xy+1)=0 ┌ WHY?
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)]
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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
最大值为 (1/2)*(4)^2-1= 7 , 最小值= -1
case2: (x,y)= +/- (1, -1)时, 代入f(x,y)得7
故f(x,y)最大值为 7 (此时 (x,y)=+/-(1, -1)), 最小值为 -1 (此时 (x,y)=+/- (1,1)
)
该怎麽从
x^2=y^2 or xy= -1 and x^2+y^2=k,
推得 (x^2, y^2)= (k/2, k/2) or [[[[[[ (x,y)=(1, -1), (-1, 1)]]]]]]
後面那一个式子呢 不是只能得到 x=1/y 然後代到 x^2+y^2=k 吗?
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