z=x²+xy+y²+x-y+1
=x²/2+xy+y²/2+x²/2+x+1/2+y²/2-y+1/2
=(1/2)(x+y)²+(1/2)(x+1)²+(1/2)(y-1)²
平方項(xiàng)恒非負(fù),z≥0,此時(shí)
x+y=0
x+1=0
y-1=0
解得x=-1 y=1
當(dāng)x=-1,y=1時(shí),函數(shù)有極小值zmin=0,沒有極大值.