已知直角三角形ABC的斜边AB=1,以AC为边长做正方形ACDE,求直角三角形ABC面积+正方形ACDE面积的最大值

题目

已知直角三角形ABC的斜边AB=1,以AC为边长做正方形ACDE,求直角三角形ABC面积+正方形ACDE面积的最大值

答案

AB=1
AC=AB×sinB=sinB
BC=AB×cosB=cosB
S△ABC+S正方形ACDE
=1/2×AC×BC+AC^2
=1/2×sinBcosB+(sinB)^2
=1/4×sin2B+(1-cos2B)/2
=1/2+5^0.5/4×(1/5^0.5×sin2B-2/5^0.5×cos2B)
=1/2+5^0.5/4×(sin(2B-arccos(1/5^0.5))
sin(2B-arccos(1/5^0.5))最大值=1
面积最大值=(2+5^0.5)/4