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

Explanation

Given, length of wire is l,
Let length of one part be x, then other part be `l-x.`
Let x part be bent into a circle and `l-x` part be in a square.
`therefore 2pir=x and 4s=l-ximpliess=(l-x)/(4) (becauses=` side)
According to question,
`r=s/2=(l-x)/(8)`
Also, `r=(2)/(2pi)`
`implies(l-x)/(8)=(x)/(2pi)`
`impliesl=(8pi)/(2pi)+x=(4x)/(pi)+x`
Sum of areas of circle and square,
`f(x)=pi((1-x)/(8))^(2)+((l-x)/(4))^(2)`
`f(x)=(pi)/(64)(l-x)^(2)+4(l-x)^(2)`
`f(x)=pi/64[(l-x)^(2)+4(l-x)^(2)]`
`f(x)=(pi)/(64)*5(l-x)^(2)`
`=(5pi)/(64)((4pi)/(pi)+x-x)^(2)`
`f(x)=(5pi)/(64)(16x^(2))/(pi^(2))`
`f(x)=(5x^(2))/(4pi)`
`therefore f'(x)=(5)/(4pi)2x=(5x)/(2pi)`
For minimum value, `f'(x)=0`
`(5)/(2pi)x=0 `
`x=0`
If `x=0,f'(0)=(5)/(2pi)gt0`
`therefore f(x)` attains minimum at `x=0.`