已知实数x、y满足x+y=1,则函数z=3x²+3y²+4xy+7的最小值是?最大值是?

y=3(x²+2xy+y²)-2xy+7
=3(x+y)²-2xy+7
=10-2xy
=10-2x(1-x)
=10+2(x²-x)
=10+2[x-(1/2)]²-(1/2)
=(19/2)+2[x-(1/2)]²
易知,在题设条件下,
(z)min=19/2
(z)max不存在