由∂f/∂x=-2x+6=0和∂f/∂y=3y²-12=0得x=3,y=2和x=3,y=-2令A=∂²f/∂x²=-2<0B=∂²f/∂x∂y=0C=∂²f/∂y²=6y在x=3,y=2处,AC-B²=-24<0,故(3,2)不是极值点;在x=3,y=-2处,AC-B²=24>0,且A<0,故(3,-2)是极大值点,极大值为f(3,-2)=30
电动车被偷了