推 free94531 :很詳細~!感謝! 07/05 22:28
※ 引述《free94531 (FreeBlizzard)》之銘言:
: 題目
: Find the point on the graph of z = x^2+y^2+10 nearest the plane x+2y-z=0?
: 答案
: (1/2,1,45/4)
: 跪求解答
: 感謝各位大大~!
非平面上之點到平面 x + 2y - z = 0 之距離
= |x+2y-z| / √(1^2+2^2+1^2) = √(x+2y-z)^2 / √(1^2+2^2+1^2)
目標函數 :√(x+2y-z)^2
限制方程式: x^2 + y^2 - z + 10 = 0
應用Lagrange Multiplier方法解目標函數之最小可能
▽{√(x+2y-z)^2} = λ.▽{x^2 + y^2 - z + 10}
=> (1,2,-1)*(x+2y-z)/√(x+2y-z)^2 = λ*(2x,2y,-1)
=> 解得 (x,y) = ( 1/2 , 1 )
=> 代回 x^2 + y^2 - z + 10 = 0,得 z = 45/4
=> ( x , y , z ) = ( 1/2 , 1 , 45/4 )
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※ 編輯: RAINDD 來自: 114.25.120.26 (07/05 22:29)