Question

利用拉格朗日乘法求f(x,y)=2(xz+yz)+xy在xyz=4的条件极值？

Answer 1

L = 2(xz+yz) + xy - k(xyz-4)
Lx = 2z+y-kyz = 0 (1)
Ly = 2z+x-kxz = 0 (2)
Lz = 2(x+y) - kxy = 0 (3)
Lk = xyz-4 = 0 (4)
(1)-(2), 得 y-x = kz(y-x), 得 y = x
代入 (3) 得 4x - kx^2 = 0 ，x = 4/k = y,
代入 (2) 得 z = -x/(2-kx) = x/2
都代入 (4)，得 x^3/2 = 4, x = 2 = y, z = 1.

Answer 2

你这明明是三个未知数,怎么是f(x,y)?
设g(x,y,z,s)=2(xz +yz)+xy +s(xyz-4),然后求
g对x,y,z,s的偏导数=0,就得到4个方程,解得到的x,y,z带入f(x,y,z)即可

Answer 3

你这明明是三个未知数,怎么是f(x,y)?设g(x,y,z,s)=2(xz yz) xy s(xyz-4),然后求g对x,y,z,s的偏导数=0,就得到4个方程,解得到的x,y,z带入f(x,y,z)即可