एक वृत्त और एक वर्ग के परिमाप का योग k है, जहां k एक अचर है सिद्ध कीजिए कि उनके क्षेत्रफल का योग निम्नतम है जब वर्ग की भुजा वृत्त की त्रिज्या की दुगनी है|

Explanation

Let r be the radius of the circle and a be the side of the square then we have
`2pir+4a=k` where k is constant
or `a=(k-2pir)/(4)`
The sum of the areas of the circle and the square (A) is given by
`A=pir^(2)+a^(2)`
`=pir^(2)+(k-2pir)/(16)^(2)`
Now `(dA)/(dr)=0`
or `2pir=(pi(k-2pir))/(4)`
or `8r=k-pir`
or `(8+2pi)r=k`
or `r=(k)/(8+2pi)=(k)/(2(4+pi))`
now , `(d^(2)4)/(dr^(2))=2pi+(pi)^(2)/(2(4+pi))`
Therefore when `r=(k)/(2(4+pi),(d^(2)A)/(dr^(2)) gt 0`
Thus the sum the of the area is least when `r= (k)/(2(4+pi))`
Thus the sum of the areas is least when `r =(k)/(2(4+pi))`
Hence it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.