Question

Sum of circumference of a circle and perimeter of a square is constant. Prove that the sum of their areas is minimtun when the ratio of the radius of the circle to a side of the square is 1:2.

Answer 1

Answer:

the ratio of the radius of the circle to a side of the square is 1:2.

Step-by-step explanation:

Let say Radius = R

& square side = A

Then 2πR + 4A  = c  (constant)

=> A = (c - 2πR)/4

Sum of Areas = S =  πR² +  A²

S = πR²   + ((c - 2πR)/4)²

dS/dR  = 2πR + (1/16) ( 2 (c - 2πR)(- 2π)

2πR  - (π/4)(c - 2πR)

dS/dR  = 0

=> 2πR = (π/4)(c - 2πR)

=> 2R = (c - 2πR)/4

=> 2R = A

R/A = 1/2

=> R  : A = 1 : 2

d²S/dR² = 2π - (π/4)(-2π) = 2π(1 + π/4)  is + ve

so Sum of Areas is minimum  when R  : A = 1 : 2

=> the ratio of the radius of the circle to a side of the square is 1:2.