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 :

  • Option 1)

    2x = (π+4) r

     

  • Option 2)

     (4−π) x = πr

     

  • Option 3)

     x = 2r

     

  • Option 4)

     2x = r

     

 

Answers (2)

As we learnt in 

Circle -

A circle is the locus of a moving point such that its distance from a fixed point is constant.

- wherein

 

 Let the length of two parts be 'a' and '2-a'

a=4x  and  2-a=2\pi r

where x is side of square

r is radius of circle

sum of Areas = \frac{a^2}{16}+\frac{a^2-4a+4}{4\pi }

of \left ( a \right ) = \frac{a^2\pi +4a^2-16a+16}{16\pi }

f^{'}\left ( a \right ) = \frac{1}{16\pi }\left [ 2a\pi +8a-16 \right ]=0

2a\pi +8a=16

x= \frac{a}{4} = \frac{2}{\pi +4}

r=\frac{2-a}{2\pi }

r=\frac{1}{\pi +4}

Hence   x=2r


Option 1)

2x = (π+4) r

 

Incorrect

Option 2)

 (4−π) x = πr

 

Incorrect

Option 3)

 x = 2r

 

Correct

Option 4)

 2x = r

 

Incorrect

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