作者LuisSantos ( )
看板trans_math
标题Re: [微分] 帮忙求一下 极值及鞍点
时间Mon Jun 9 20:59:50 2008
※ 引述《xxoo225 (xxoo)》之铭言：
: f(x,y)=8x*3+y*3+6xy
: 求其极值和鞍点??
: *後面的是次方数
: 谢谢啦~!!
f(x,y) = (8)(x^3) + y^3 + 6xy
fx = (24)(x^2) + 6y = 0 ------(1)
fy = (3)(y^2) + 6x = 0 ------(2)
(1) => y = (-4)(x^2) 代入(2) 得
(3)(16)(x^4) + 6x = 0
(8)(x^4) + x = 0 => (x)((8)(x^3) + 1) = 0
x = 0 或 (8)(x^3) + 1 = 0
x = 0 或 x^3 = -1/8 => x = 0 或 -1/2
当 x = 0 时 , y = 0
当 x = -1/2 时 , y = (-4)(1/4) = -1
所以(0,0)和(-1/2 , -1)为临界点
fxx = 48x , fxy = 6 = fyx , fyy = 6y
D(x,y) = (fxx)(fyy) - (fxy)^2
= (48x)(6y) - 36 = 288xy - 36
D(0,0) = -36 < 0 => (0,0) 为鞍点
D(-1/2 , -1) = (288)(-1/2)(-1) - 36 = 144 - 36 = 108 > 0
因为 fxx(-1/2 , -1) = -24 < 0
所以当(x,y) = (-1/2 , -1) 时 ,
f(x,y) = f(-1/2 , -1)
= (8)(-1/8) - 1 + (6)(-1/2)(-1)
= -1 - 1 + 3 = 1 为相对极大值
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