解(拉格朗日乘数法):设F=xy+λ(x+y-1)
令Fx=y+λ=0........(1)
Fy=x+λ=0........(2)
Fλ=x+y-1=0........(3)
解方程组(1),(2)得x=y
代入(3)得x=y=1/2
故z的极大值=z(1/2,1/2)=(1/2)(1/2)=1/4
解(拉格朗日乘数法):设F=xy+λ(x+y-1)
令Fx=y+λ=0........(1)
Fy=x+λ=0........(2)
Fλ=x+y-1=0........(3)
解方程组(1),(2)得x=y
代入(3)得x=y=1/2
故z的极大值=z(1/2,1/2)=(1/2)(1/2)=1/4