Find the minimum value of
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Find the minimum value of
2014-04-30 8:18 pm
回答 (10)
2014-04-30 9:14 pm
✔ 最佳答案
Sol2x+y=1
y=1－2x
A=x^2+xy+y^2
=x^2+x(1－2x)+(1－2x)^2
=x^2－2x^2+x+4x^2－4x+1
=3x^2－3x+1
=3(x^2－x)+1
=3(x^2－x+1/4)+1/4
=3(x－1/2)^2+1/4
When x=1/2
Min=1/4
2014-06-05 2:52 pm
2014-05-30 4:40 pm
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2014-05-01 11:44 am
2014-04-30 11:16 pm
微積分的作法(Lagrange 乘子法)
Let L(x,y)=x^2 +xy +y^2 +λ(2x+y-1)
∂L /∂x = 2x+y+2λ=0.....(1)
∂L /∂y = x+2y+λ=0....(2)
由(2),λ= -x-2y 代入(1)
得 y=0 代入 2x+y=1 得 x=1/2
x=1/2, y=0 代入x^2 +xy + y^2 得最小值 1/4
Let L(x,y)=x^2 +xy +y^2 +λ(2x+y-1)
∂L /∂x = 2x+y+2λ=0.....(1)
∂L /∂y = x+2y+λ=0....(2)
由(2),λ= -x-2y 代入(1)
得 y=0 代入 2x+y=1 得 x=1/2
x=1/2, y=0 代入x^2 +xy + y^2 得最小值 1/4
收錄日期: 2021-05-02 10:46:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140430000010KK02952