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) \[\therefore \]\[x=2r\]