Question

高等数学三个题目。求过程

Answer 1

4. 记 u=yz, v=e^(xz), 则 F(u, v)=0,
F'<x> = F'<u>u'<x>+F'<v>v'<x> = ze^(xz)F'<v>,
F'<y> = F'<u>u'<y>+F'<v>v'<y> = zF'<u>,
F'<z> = F'<u>u'<z>+F'<v>v'<z> = yF'<u>+xe^(xz)F'<v>,
z'<x> = -F'<x>/F'<z> = -ze^(xz)F'<v>/[yF'<u>+xe^(xz)F'<v>],
z'<y> = -F'<y>/F'<z> = - zF'<u>/[yF'<u>+xe^(xz)F'<v>],
dz = z'<x>dx+z'<y>dy = -z[e^(xz)F'<v>dx+F'<u>dy]/[yF'<u>+xe^(xz)F'<v>],

5. f(x,y)=4(x-y)-x^2-y^2, f'<x>=4-2x, f'<y>=-4-2y, 联立解得驻点 (2，-2).
f''<x>=-2<0, f''<xy>=0, f''<yy>=-2, AC-B^2=(-2)(-2)-0^2=4>0,
故点 (2，-2) 为极大值点， 极大值 f(x,y)=8.

6. 构造拉格朗日函数 L=x-2y+2z+λ(x^2+y^2+z^2-1),
L'<x>=0: 1+2λx=0;
L'<y>=0: -2+2λy=0;
L'<z>=0: 2+2λz=0;
L'<λ>=0: x^2+y^2+z^2=1.
前三式得，x=-1/(2λ), y=1/λ, z=-1/λ, 代人x^2+y^2+z^2=1，
得 λ=±3/2, 则条件极值点 A(-1/3, 2/3, -2/3), B(1/3, -2/3, 2/3),
u(-1/3, 2/3, -2/3)=-3, u(1/3, -2/3, 2/3)=3,