Re: [多變] 問極值?
※ 引述《chat543 (小ㄐ)》之銘言：
: Find the maximum and minimum values of f(x,y,z) =x-2y+z , with the
: constraints x^2+y^2+z^2 = 1 and x+y+z=0.
: 他有兩個限制條件~不知道要怎嚜求?
令 f(x,y,z) = x - 2y + z
g1(x,y,z) = x^2 + y^2 + z^2 - 1
g2(x,y,z) = x + y + z
F(x,y,z) = f(x,y,z) + (λ1)(g1(x,y,z)) + (λ2)(g2(x,y,z))
則 F(x,y,z) = f(x,y,z) + (λ1)(g1(x,y,z)) + (λ2)(g2(x,y,z))
= x - 2y + z + (λ1)(x^2 + y^2 + z^2 - 1) + (λ2)(x + y + z)
Fx = 1 + (λ1)(2x) + λ2 = 0 ------(1)
Fy = -2 + (λ1)(2y) + λ2 = 0 ------(2)
Fz = 1 + (λ1)(2z) + λ2 = 0 ------(3)
(1) - (3) => (λ1)(2x) - (λ1)(2z) = 0 => (2)(λ1)(x - z) = 0
λ1 = 0 或 x = z
如果 λ1 = 0
代入(1) 1 + λ2 = 0 => λ2 = 1
代入(2) -2 + λ2 = 0 => λ2 = 2
因此 λ1 = 0 不合
所以 x = z
x = z 代入 x^2 + y^2 + z^2 = 1 和 x + y + z = 0 得
(2)(x^2) + y^2 = 1 ------(4)
2x + y = 0 ------(5)
(5) => y = -2x 代入(4)得 (2)(x^2) + (4)(x^2) = 1 => (6)(x^2) = 1
1 -2 1 -1 2 -1
(x,y,z) = (----- , ----- , -----) 、 (----- , ----- , -----)
√6 √6 √6 √6 √6 √6
1 -2 1
當 (x,y,z) = (----- , ----- , -----) 時 ,
√6 √6 √6
1 -2 1
f(x,y,z) = f(----- , ----- , -----)
√6 √6 √6
1 4 1 6
= ----- + ----- + ----- = ----- = √6
√6 √6 √6 √6
-1 2 -1
當 (x,y,z) = (----- , ----- , -----) 時 ,
√6 √6 √6
-1 2 -1
f(x,y,z) = f(----- , ----- , -----)
√6 √6 √6
-1 4 1 -6
= ----- - ----- - ----- = ----- = -√6
√6 √6 √6 √6
1 -2 -1
因此當(x,y,z) = (----- , ----- , -----) 時 , f(x,y,z) 有極大值 √6
√6 √6 √6
-1 2 -1
當(x,y,z) = (----- , ----- , -----) 時 , f(x,y,z) 有極小值 -√6
√6 √6 √6
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