Question

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.

Answer 1

Consider x,y as the radius of the circle and side of the square respectively.

Write the equation for the given constant k.

2πx+4y=k 4y=k−2πx y= k−2πx 4 (1)

Consider z as the sum of areas of circle and square.

Write the equation for the sum of areas of circle and square,

z=π x 2 + y 2

Substitute the value from equation (1)

z=π x 2 + ( k−2πx 4 ) 2 = 16π x 2 + k 2 +4 π 2 x 2 −4kπx 16 = 1 16 [ ( 16π+4 π 2 ) x 2 −4kπx+ k 2 ]

Differentiate both sides of the equation,

dz dx = 1 16 [ ( 16π+4 π 2 )2x−4kπ ] d 2 z d x 2 = 1 16 [ ( 16π+4 π 2 )2 ]

Now,

dz dx =0 1 16 [ ( 16π+4 π 2 )2x−4kπ ]=0 4π( 4+π )2x=4kπ x= k 2( 4+π )

At, x= k 2( 4+π ) ,

d 2 z d x 2 = 1 16 ( 16π+4 π 2 )2 >0

Thus, z is minimum at x= k 2( 4+π )

From equation (1),

y= k−2π( k 2( 4+π ) ) 4 = 4k+πk−πk 4( 4+π ) =2× 4k 2( 4+π ) =2x

Thus, sum of the areas is minimum when side of the square is double the radius of the circle.