How do you prove that if the sum of the radii of two circles remains constant, the sum of the areas of the circles is least when the circles are equal?

Answer 1

R+r = k

pi*R^2 + pi*r^2 = pi * (k-r)^2 + pi*r^2

= pi*(k^2 - 2kr + r^2) + pi*r^2

= pi * k^2 - 2* k * pi * r + 2 * pi * r^2

this is a quadratic function in terms of r

with A = 2*pi , B = -2 * k * pi, and C = pi * k^2

THe extrema occurs at -B/(2a)

-B/(2A) = 2*k*pi/ (4*pi) = k/2

so when r=k/2, the R + r = k

R + k/2 = k

R = k/2

therefore the areas are the same when r=R=k/2