作者LuisSantos ( )
看板trans_math
标题Re: [单变] 请问一题计算题
时间Fri Jul 4 18:05:44 2008
※ 引述《flygey (努力达成目标)》之铭言：
: The point no the ellipse 4x^2+9y^2=36,that is nearest the origin is ________
: 请问版友这题如何算
: 感谢
令 f(x,y) = x^2 + y^2 , g(x,y) = 4x^2 + 9y^2 - 36
F(x,y) = f(x,y) + (λ)(g(x,y))
= x^2 + y^2 + (λ)(4x^2 + 9y^2 - 36)
Fx = 2x + (λ)(8x) = 0 => (2x)(1 + 4λ) = 0
-1
=> x = 0 或 λ = --- ------(1)
4
Fy = 2y + (λ)(18y) = 0 => (2y)(1 + 9λ) = 0
-1
=> y = 0 或 λ = --- ------(2)
9
由(1)和(2)得
-1 -1
λ = --- 且 λ = --- 明显不合
4 9
(x,y) = (0 , 2) , (0 , -2) , (3 , 0) , (-3 , 0)
f(0,2) = 4
f(0,-2) = 4
f(3,0) = 9
f(-3,0) = 9
所以当(x,y) = (0,2) , (0,-2) 时 ,
f(x,y)有最小值4 , 即离原点最短距离为2
另解
x^2 y^2
4x^2 + 9y^2 = 36 => ----- + ----- = 1
9 4
画xy座标图得知
x^2 y^2
在椭圆 ----- + ----- = 1 上离原点最近的点为 (0,2)和(0,-2)
9 4
因此最短距离为 2
--
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