Question

The sum of the perimeter of a circle and 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 the circle.

Answer 1

Let r be the radius of the circle and x be the side of the square, then

2πr + 4x = k                        ....(i)

Let S be the sum of areas of the circle and the square, then

`S = pir^2 + x^2 = pir^2 + ((k - 2pir)/4)^2`

`("from" (i), x = (k-2pir)/4)`

`pir^2 + k^2/16 + (pi^2r^2)/4  - (kpir)/4`              ....(ii)

Differentiating (ii) w.r.t x, we get

`(dS)/(dr) = 2pir + (2pi^2r)/4 - (kpi)/4`               .....(iii)

For maximum / minimum, let `(dS)/(dr) = 0`

⇒ `2pir + (2pi^2r)/4 - (kpi)/4 = 0`

⇒` r (2pi + pi^2/2) = (kpi)/4`

⇒ `r (2kpi)/ (4(4 pi + pi^2)) = k/(2 (4 + pi))`

Differentiating (iii) w.r.t x, we get,

`(d^2S)/(dr^2) = 2pi + pi^2/2 > 0   AAr`

`((d^2S)/(dr^2))_(r = k/(2(4 + pi))) > 0`

⇒ S is minimum at `r = k/(2(4 + pi)).`

⇒ `x = 1/4{k - (2pi)/2 (k/ (4 + pi))}`

`= (4k)/(4 (4 + pi)) = 2 [k/ (2 (4 + pi))] = 2 (r).`

Hence, S is least when side of the square is double the radius of the circle.