Asked by Adeeb Khan | 22nd Jul, 2014, 06:40: PM

Expert Answer:

G i v e n space t h a t space t h e space p e r i m e t e r space o f space a space c i r c l e space a n d space a space c i r c l e space a n d space a space s q u a r e space i s space K K equals 2 πr plus 4 straight a rightwards double arrow 4 straight a equals straight K minus 2 πr rightwards double arrow straight a equals fraction numerator straight K minus 2 πr over denominator 4 end fraction.... left parenthesis 1 right parenthesis The space sum space of space the space areas space of space the space circle space and space square space is space given space by straight A equals πr squared plus straight a squared rightwards double arrow straight A equals πr squared plus open parentheses fraction numerator straight K minus 2 πr over denominator 4 end fraction close parentheses squared fraction numerator d straight A over denominator d straight r end fraction equals 2 πr plus 2 open parentheses fraction numerator straight K minus 2 πr over denominator 16 end fraction close parentheses open parentheses minus 2 straight pi close parentheses rightwards double arrow fraction numerator d straight A over denominator d straight r end fraction equals 2 πr minus straight pi open parentheses fraction numerator straight K minus 2 πr over denominator 4 end fraction close parentheses For space minimum comma space fraction numerator d straight A over denominator d straight r end fraction equals 0 2 πr minus straight pi open parentheses fraction numerator straight K minus 2 πr over denominator 4 end fraction close parentheses equals 0 rightwards double arrow 2 straight r equals fraction numerator straight K minus 2 πr over denominator 4 end fraction rightwards double arrow 8 straight r equals straight K minus 2 πr rightwards double arrow 8 straight r plus 2 πr equals straight K rightwards double arrow straight r open parentheses 8 plus 2 straight pi close parentheses equals straight K rightwards double arrow straight r equals fraction numerator straight K over denominator open parentheses 8 plus 2 straight pi close parentheses end fraction.... left parenthesis 2 right parenthesis Now space let space us space find comma space fraction numerator d squared straight A over denominator d straight r squared end fraction space fraction numerator d squared straight A over denominator d straight r squared end fraction equals 2 straight pi plus straight pi over 2 greater than 0 Thus comma space area space is space minimum space when comma space straight r equals fraction numerator straight K over denominator open parentheses 8 plus 2 straight pi close parentheses end fraction straight a equals fraction numerator straight K minus 2 straight pi cross times fraction numerator straight K over denominator open parentheses 8 plus 2 straight pi close parentheses end fraction over denominator 4 end fraction rightwards double arrow a equals fraction numerator straight K minus straight pi cross times fraction numerator straight K over denominator open parentheses 4 plus straight pi close parentheses end fraction over denominator 4 end fraction rightwards double arrow a equals fraction numerator straight K open parentheses 4 plus straight pi close parentheses minus K straight pi over denominator 4 open parentheses 4 plus straight pi close parentheses end fraction rightwards double arrow a equals fraction numerator 4 straight K over denominator 4 open parentheses 4 plus straight pi close parentheses end fraction rightwards double arrow a equals fraction numerator straight K over denominator open parentheses 4 plus straight pi close parentheses end fraction.... left parenthesis 3 right parenthesis F r o m space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e straight a equals fraction numerator straight K minus 2 πr over denominator 4 end fraction rightwards double arrow 4 a equals K minus 2 πr rightwards double arrow 2 πr equals straight K minus 4 straight a rightwards double arrow straight r equals fraction numerator straight K minus 4 straight a over denominator 2 straight pi end fraction.... left parenthesis 4 right parenthesis Substituting space the space value space of space straight a space from space equation space left parenthesis 3 right parenthesis space in space equation space left parenthesis 4 right parenthesis comma space we space have comma straight r equals fraction numerator straight K minus 4 cross times fraction numerator straight K over denominator 4 plus straight pi end fraction over denominator 2 straight pi end fraction straight r equals fraction numerator straight K over denominator 2 open parentheses 4 plus straight pi close parentheses end fraction... left parenthesis 5 right parenthesis From space equations space left parenthesis 3 right parenthesis space and space left parenthesis 5 right parenthesis comma space it space is space clear space that space a equals 2 r

Answered by Vimala Ramamurthy | 23rd Jul, 2014, 11:29: AM