Question

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.

Answer 1

let side of square = a
radius of circle = r
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, $\frac{dS}{dr} =0$

⇒$\frac{d}{dr} ( \pi r^{2}+ \frac{ (K-2 \pi r)^{2} }{16} )=0$

⇒$\frac{d}{dr} ( \pi r^{2})+ \frac{d}{dr}( \frac{ (K-2 \pi r)^{2} }{16} )=0$

⇒$2 \pi r+ \frac{1}{16}(8 \pi ^{2} r-4 \pi K)=0$

⇒$2 \pi r+ \frac{1}{16}(8 \pi ^{2} r-4 \pi (4a+2 \pi r))=0$

⇒$2 \pi r+ \frac{1}{16}(8 \pi ^{2} r-16 \pi a-8 \pi^{2} r))=0$

⇒$2 \pi r+ \frac{1}{16}(-16 \pi a)=0$

⇒$2 \pi r- \pi a=0$

⇒$2r=a$  (proved)