# A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square. - Mathematics and Statistics

A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square.

#### Solution

Length of the wire is 'l'.
Let the part bent to make circle is of length 'x',
and the part bent to make square is of length 'l - x'.
Circumference of the circle = 2πr = x

r=x/(2pi)

Area of the circle =pir^2=pi(x/(2pi))^2=x^2/(4pi)

Perimeter of the square = 4a=l-x rArra=(l-x)/4

Area of the square = ((1-x)/4)^2=(1-x)^2/16

Sum of the areas A(x) =  x^2/(4pi)+(l-x)^2/16

For extrema, (dAx)/(dx)=0

(2x)/(4pi)+(2(l-x)(-1))/16=0

(4(2x)+2pi(x-l))/16=0

4x+pix-pil=0

x=(pil)/(4+pi)

Since there is one point of extremum, it has to be the minimum in this case.

r=x/(2pi)=l/(2(4+pi)).........(1)

Side of the square a=(l-x)/4=(l-(pil)/(4+pi))/4=l/(4+pi)..................(2)

From (1) and (2), we get that the radius of the circle is half the side of the square, for least sum of areas. (Proved)

Concept: Maxima and Minima - Introduction of Extrema and Extreme Values
Is there an error in this question or solution?