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