A wire of length 2 units is cut into two parts which are bent respectively to from a square ofside c units and a circle of radius r units if the sum of the sum of the areas of the square and the circle so fromed is minimum then

Options

A. `2x=(pi 4)r`
B. `(4-pi)x=pir`
C. x=2r
D. 2x=r

Answer

C

Explanation

Let A be the sum of the areas of the square and the circle
`A =x^2+pi r^2,` whhere `4x + 2 pi r -2`
`rArr A= ((1-pir )/(2))^2+pi r^2`
`rArr A =1/4 (1- pi r)^2`
`rArr A= (1-pi r)^2+pir^2`
`rArr A= 1/4 (1-pi)^2+pir^2`
`rArr (dA)/(dr)=-pi/2 (1- pi r)+2pi r and (d^2A)/(dr^2)=(pi^(2))/(2)+2pi`
For maximum of minimum values of A, we must have
`(dA)/(dr)=rArr 1/2(1-pir)=2r rArr 4r =1 pi r rArr r =(1)/(pi +4)`
Clearly `(d^2A)/(dr^2) gt 0 ` for all r So, A is minimum when `r=1/(pi+4)`
Putting `r=(1)/(x+4) is 4x + (2 pi )/(pi +4)rArr x= 2/(pi +4)`
Clearly ,x= 2r