将长度为1的铁丝分成两段,围成正方形和圆形,要使正方形与圆形的面积之和最小,则正方形的周长为多少?

1个回答

  • 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)