用拉格朗日乘数法求下列条件极值的可疑极值点,并用无条件极值的方法确定是否取得极值:
(1)目标函数z=xy,约束条件x+y=1;
(2)目标函数z=x2+y2,约束条件x/a+y/b=1;
(3)目标函数u=x-2y+2z,约束条件x2+y2+z2=1.
【正确答案】:(1)L(x,y)=xy+λ(x+y-1) 解方程组: {Lx=y+λ=0 Ly=x+λ=0 x+y-1=0 得: x=1/2 y=1/2 λ=-(1/2) 经判定:(1/2,1/2)为极小值点,z(1/2,1/2)=1/4 (2)L(x,y)=x2+y2+λ(x/a+y/b-1) 解方程组: {Lxyλ2/(a2+b2) y=a2b2/(a2+b2) λ=-[2a2b2/(a2+b2)] 经判定[ab2/(a2+b2),a2b/(a2+b2)]为极小值点 极小值z[ab2/(a2+b2),a2b/(a2+b2)]= a2b2/(a2+b2) (3)L(X,y,z)=x-2y+2z+λ(x2+y2+z2-1) 解方程组: {Lx=1-2xλ=0 Ly=-2+2yλ=0 Lz=2+2zλ=0 Lλ=x2+y2+z2-1=0 得: x=1/3 y=-(2/3) z=2/3 λ=3/2 或 x=-1/3 y=2/3 z=-(2/3) λ=-(3/2) 经判定(1/3,-(2/3),2/3)为极大值点,(-(1/3),(2/3),-(2/3))为极小 值点极大值u(1/3,-(2/3),2/3)=3,极小值u(-(1/3),2/3,-(1/3))=-3
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