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)