Re: [微分] 幫忙求一下 極值及鞍點

看板trans_math作者 ( )時間17年前 (2008/06/09 20:59), 編輯推噓0(000)
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※ 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 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.119.29.34
文章代碼(AID): #18JIb7fO (trans_math)
文章代碼(AID): #18JIb7fO (trans_math)