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) \[\therefore \]\[x=2r\]You need to login to perform this action.
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