Question

x²+4y²+x+2y=1,则xy最大值为？

答案是4-2×2½，只求过程

Answer 1

解：
x²+4y²+x+2y=1
(x²+x+¼)+(4y²+2y+¼)=3/2
(x+½)²+(2y+½)²=3/2
令x+½=(√6/2)cosα，2y+½=(√6/2)sinα
则x=(√6cosα-1)/2，y=(√6sinα-1)/4
xy=[(√6cosα-1)/2][(√6sinα-1)/4]
=[6sinαcosα-√6(sinα+cosα)+1]/8
=[3sin(2α)-2√3cos(α-π/4)]/8
=[3cos(π/2 -2α)-2√3cos(α-π/4)]/8
=[3cos2(α-π/4)-2√3cos(α-π/4)]/8
=[3(2cos²(α-π/4)-1)-2√3cos(α-π/4)]/8
=[6cos²(α-π/4)-2√3cos(α-π/4)-3]/8
=¾[(cos(α-π/4)- √3/6)² -5/12]
-1≤cos(α-π/4)≤1
cos(α-π/4)=-1时，xy有最大值
(xy)max=[6·(-1)²-2√3·(-1)-3]/8=(3+2√3)/8
xy的最大值为(3+2√3)/8