Question

21求过程

Answer 1

21、解：BC=sinα，OB=cosα

              OA=DA÷tan(π/3)

                   =BC÷tan(π/3)

                   =sinα÷tan(π/3)

                   =(√3sinα)/3

               AB=OB-OA=cosα-(√3sinα)/3

               则矩形面积 y=AB·BC

                                    =[cosα-(√3sinα)/3]sinα

                                    =-√3/3sin²α+sinαcosα

                                    =-√3/3sin²α+sinαcosα

        

                 

        

                对于关于sinα的曲线y=-√3/3sin²α+sinαcosα而言：a=-√3/3，b=cosα，c=0

                所以当sinα=-b/a=(√3/3)cosα时，y有最大值。

                       sinα=(√3/3)cosα

                       tanα=√3/3

                       α=π/6

                 (4ac-b²)/4a=[0-cos²(π/6)]/4(-√3/3)

                                   =√3/6