Question

. The sum of the perimeter of a circle and a square is K, where K is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.(2010Sp)(2010)

Answer 1

this may help u.......

Answer 2

Hey !!

Let side of square be a units and radius of circle be r units.

It is given that 4a + 2πr = k , where k is a constant

    =>      r = k - 4a / 2π

 Sum of areas, A = a² + πr²

=> A = a² + π[ k - 4a / 2π ]² = a² + 1/4π (k - 4a)²

Differentiating w.r.t 'a'

        dA/da = 2a + 1/4π. 2(k - 4a).(-4) = 2a - 2(k - 4a) / π ------------------> (1)

For minimum area, dA/da = 0

              => 2a - 2(k - 4a) / π = 0

              => 2a = 2(k - 4a) / π

              => 2a = 2(2πr)/π                      [ As k = 4a + 2πr given ]

Now, differentiating equation (1) w.r.t to 'a'

d²A/da² = 2 - 2/π(-4) = 2 + 8/π

at a = 2π , d²A/da² = 2 + 8/π > 0

∴ For ax = 2r, sum of areas is least

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

GOOD LUCK !!