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.

Explanation

Let x be the side of the square and r be the radius of the circle.
According to the problem
Let P = perimeter of square+ circumference of circle
` rArr P = 4x + 2 pi r` ….(1)
Let `A = pi r^(2) + x^(2)`
` = pi r^(2) = ((P-2 pi r )/4)^(2)` [ From eqn. (1)]
` = pi r^(2) + 1/16 (P-2pi r)^(2)`
` rArr (dA)/(dr) = 2 pi r +2/16 (-2pi)(p-2 pi r)`
` = 2 pi r - pi/4 (P-2 pi r)`
and ` (d^(2)A)/(dr^(2)) = 2 pi + pi^(2)/2`
From maxima/minima
` (dA)/(dr) = 0 `
` 2 pi r - pi/4 (P-2 pi r) =0` ...(2)
` rArr 2 pi r = pi/4 4 x`
` rArr 2 r = x `
at x = 2r,
` (d^(2)A)/(dr^(2)) gt 0 `
` rArr` A is minimum.
Therefore, total area will be minimum if the side of the square is equal to the diameter of the circle.