A wire of length 2 units is cut into two parts which are bent respectively to form a square of `s i d e=x` units and a circle of `r a d i u s=r` units. If the sum of the areas of the square and the circle so formed is minimum, then : (1) `2x=(pi+4)r` (2) `(pi+4)x=pir` (3) `x=2r` (4) `2x=r`

Explanation

`4x + 2 pi r = 2`

`2x + pi r = 1`

`r = (1 - 2 x)/pi`

`f(x)=`area os square of side x + area of circle with radius r

`f(x) = x^2 + pi r^2 `

`f(x) = x^2 + pi ((1- 2x)/pi)^2`

`f'(x) = 0`

`2x + pi xx 2 ((1-2x)/pi)xx ((0-2)/pi)=0`

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