question_answer

A wire of length 2 units is cut into two parts  which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then : [JEE Main Solved Paper-2016 ]

A) \[2x=r\]                               

B) \[2x=(\pi +4)r\]

C) \[(4-\pi )x=\pi r\]            

D) \[x=2r\]

Correct Answer: D

Solution :

given that \[4x+2\pi r=2\] i.e. \[2x+\pi r=1\] \[\therefore \]\[r=\frac{1-2x}{\pi }\]                                                      ..... (i) Area \[={{x}^{2}}+\frac{1}{\pi }{{(2x-1)}^{2}}\] \[\frac{dA}{dx}=0\]gives\[x=\frac{2}{\pi +4}\]for min value of area A from (i) & (ii) \[r=\frac{1}{\pi +4}\]                                      ..... (iii) \[\therefore \]\[x=2r\]