推 abbybao :感謝解題:) 11/20 13:05
※ 引述《abbybao (小寶)》之銘言：
: 想問一下如果要用Lagrange法做要怎麼做?
: 我把式子列出來後不知道怎麼化簡
: http://ppt.cc/WJcs
2x = λ(2x+y-3) ...(1)
2y = λ(x+2y-3) ...(2)
x^2+xy+y^2-3x-3y-9 = 0 ....(3)
若2x+y-3 = 0 代入(1) 得 x = 0, y = 3
但代入(3)不滿足 故 2x+y-3 ≠ 0
同理 x+2y-3 ≠ 0
由(1)(2)得 λ = 2x/(2x+y-3) = 2y/(x+2y-3)
交叉相乘 x^2+2xy-3x = y^2+2xy-3y
(x^2-3x+9/4) = (y^2-3y+9/4)
(x-3/2)^2 = (y-3/2)^2
x-3/2 = y-3/2 or x-3/2 = -y+3/2
=> x = y or x = -y+3
(i) x = y 代入(3)
3y^2-6y-9 = 0 => y^2-2y-3 = 0 => (y+1)(y-3) = 0, y = -1 or y = 3
(ii) x = -y+3 代入(3)
(-y+3)^2 +(-y+3)*y+y^2-3(-y+3)-3y-9 = 0
=> y^2-3*y-9 = 0 => y = 1.5*(1+√5) or y = 1.5*(1-√5)
共4組解 (x,y) = (-1,-1), (3,3), (1.5*(1-√5),1.5*(1+√5)),
(1.5*(1+√5),1.5*(1-√5))
x^2+y^2 = 2, 18, 27, 27
最小值 = 2, 最大值 = 27
: 另外再問一題
: http://ppt.cc/FAHa
: z2的主幅角不能是10度嗎?
: 這樣也能讓z1*z2在虛軸上
: 拜託大家了,謝謝^^
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