设圆的周长为x则正方形周长为（1-x）S＝π （x/2π）^2 +[(1-x)/4]^2 ＝x^2/4π +（x-1）^2/16 ＝[（16+4π)/64π][(x^2-8πx/(16+4π）+4π/(16+4π)] ＝[（16+4π)/64π][(x-4π/（16+4π)]^2+1/16-π/（64+16π）...