# The sum of perimeter of a circle & square is K.where K is some constant.Prove that the sum of their areas is least when the side of square is double the radius of circle.

2
by Raman786
i need detail

2014-10-24T22:41:05+05:30

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Let side of square = a
perimeter of square = 4a
perimeter of circle = 2πr
sum = 4a+2πr = K
⇒4a = K-2πr
⇒a = (K-2πr)/4

area of square = a² = ((K-2πr)/4)² = (K-2πr)²/16
area of circle = πr²
sum of areas, S = πr² + (K-2πr)²/16

For minimum area,

(proved)

very thank ful 2 u
:)
2014-10-25T12:50:02+05:30
P(circle) =2pie r
p(square)=4s
a.t.q.
2pie r+4a=k
pie r2 +a2
(given that) 2r=a
assuming the value of r=1
then, 2*22/7*1+8(1)=100/7 (sum of peri. )
22/7*(1)square+4(1)square=50/7
so here area is minimum