Question icon
Grade 12Algebra

1. 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:

Profile image of Narendra
8 Years agoGrade 12
Answers icon

1 Answer

Profile image of Samyak Jain
8 Years ago

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.

Perimeter of square = 4x  &  perimeter of circle (i.e. its circumference) = 2\pir
So sum of perimeters of given figures is equal to 2,
i.e., 4x + 2\pir = 2  or  x = (1 – \pir)/2
Sum of areas of the square & the circle = x\pir2 , Put x = (1 – \pir)/2
= [(1 – \pir)/2]\pir2
= (1 – 2\pir + \pi2 r2)/4 + \pir2
= (1/4)[1 – 2\pir + \pi2 r+4\pir2]
Differentiate the above expression wrt x and equate it to zero.
\therefore (1/4)[0 – 2\pi + \pi2 2r + 4\pi.2r] = 0  => – 2\pi + \pi2 2r + 4\pi.2r = 0
2(\pi+ 4\pi) r = 2\pi  => r = \pi / (\pi+ 4\pi) = 1/(\pi + 4)
Differentiate (1/4)[0 – 2\pi + \pi2 2r + 4\pi.2r] to check the value of r we got is minimum or maximum.
We get (1/4)[0 – 0 + 2\pi2 + 8\pi] > 0. Thus, minima is formed.
\because x = (1 – \pir)/2 ,   x = (1 – \pi/\pi + 4)/2  = (\pi + 4 – \pi) / 2(\pi + 4)
x = 4/2(\pi + 4) = 2 / \pi + 4
\therefore x = 2 / (\pi + 4) & r = 1/(\pi + 4) .
If satisfied with my answer, pls approve it.