Question

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Answer 1

Answer:

Let  the length of pieces of wire are x & 28-x

consider the perimeter of square(side=a) =x & circumference of circle(radius=r)=28-x

   so, 4a=x

         a=x/4

and, 2*pi*r=(28-x)

            r=(28-x)/2*pi             

  combine area (A) = area of square+area of circle  

                            A =  [{(x/4)^2} + {pi*(28-x/2*pi)^2}]  

                          when we solved this equestion ,we get,

     A = x^2/16 + (28^2)/(4*pi) + x^2/(4*pi) - 14x/pi  --------------------------eqn(1)

A will be minimum ,when dA/dx=0

dA/dx = 2x/16 + 0 + 2x/4pi - 14/pi

  0       = x/8 + x/2pi - 14/pi

  14/pi =x/2(1/4+1/pi) 

  (14*2*4pi)/pi*(pi+4) = x

    x = 112/(pi+4)

    x= 15.68 metre               when,pi=22/7

  put the value of x in eqn(1)

   A  (minimum) = 38.12 metre^2

Answer 2

Step-by-step explanation:

38.12 meter is your answer

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