作者LuisSantos ( )
看板trans_math
標題Re: [微分] 幫忙求一下 極值及鞍點
時間Mon Jun 9 20:59:50 2008
※ 引述《xxoo225 (xxoo)》之銘言：
: f(x,y)=8x*3+y*3+6xy
: 求其極值和鞍點??
: *後面的是次方數
: 謝謝啦~!!
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
當 x = 0 時 , y = 0
當 x = -1/2 時 , y = (-4)(1/4) = -1
所以(0,0)和(-1/2 , -1)為臨界點
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) 為鞍點
D(-1/2 , -1) = (288)(-1/2)(-1) - 36 = 144 - 36 = 108 > 0
因為 fxx(-1/2 , -1) = -24 < 0
所以當(x,y) = (-1/2 , -1) 時 ,
f(x,y) = f(-1/2 , -1)
= (8)(-1/8) - 1 + (6)(-1/2)(-1)
= -1 - 1 + 3 = 1 為相對極大值
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