答案是80/(4+π)
设圆周长为C,则正方形周长为20-C
总面积S=C^2/(4π)+[(20-C)/4]^2=[(4+π)C^2-40πC+400π]/(16π)
当C=-b/(2a)=20π/(4+π)时S取最小值
此时
正方形的周长为20-C=20-20π/(4+π)=80/(4+π)
答案是80/(4+π)
答案是80/(4+π)
设圆周长为C,则正方形周长为20-C
总面积S=C^2/(4π)+[(20-C)/4]^2=[(4+π)C^2-40πC+400π]/(16π)
当C=-b/(2a)=20π/(4+π)时S取最小值
此时
正方形的周长为20-C=20-20π/(4+π)=80/(4+π)
答案是80/(4+π)