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)
当x=π/(π+4)时 ,面积最小
正方形周长=1-π/(π+4)
=4/(π+4)
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)
当x=π/(π+4)时 ,面积最小
正方形周长=1-π/(π+4)
=4/(π+4)