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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.
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Solution
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.
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