由y=(sinx+1)(cosx+1)→y=(sinx+cosx)/2 + (sinx+cosx) + 1/2→y=[√2sin(x+π/4)+1]/2 由x∈[-π/6,π/2]→π/12≤x+π/4≤3π/4→y的最大值为3/2 + √2.