. 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)
Answers
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 !!