Question

A wire of length \[l\]is cut into two parts. One part is bent into a circle and another into a square. Show that the sum of the areas of the circle and square is the least, if the radius of the circle is half the side of the square.

Answer 1

Hint:
The circle has Circumference is \[2\pi r\] and square has \[4s\] Circumference where ‘\[r\]’ is radius and S is Side of Square.
Let the x part be bent in a circle and \[l - x\] part be in square.

 Complete step by step solution:
Given, Length of wire is ‘L’
Let Length of one part be \[x\].
Then the other part is \[l - x\].
\[\therefore \] Circumference of Circle \[ \Rightarrow x = 2\pi r\] and square \[4s = l - x\].
  \[s = \dfrac{{l - x}}{4}\].
S is Side of Square.
According to the question we have,
\[r = \dfrac{s}{2} = \dfrac{{l - x}}{8}\]
Also, \[r = \dfrac{x}{{2\pi }}\]
\[\dfrac{{l - x}}{8} = \dfrac{x}{{2a}}\]
\[ \Rightarrow l = \dfrac{{8x}}{{2\pi }} + x = \dfrac{{4x}}{\pi } + x\]
Now, Sum of Areas of Circle and Square.
  \[f(x) = \pi {\left( {\dfrac{{l - x}}{8}} \right)^2} + {\left( {\dfrac{{l - x}}{4}} \right)^2}\]
\[f(x) = \dfrac{\pi }{{64}}[{(l - x)^2} + 4{(l - x)^2}]\]
\[f(x) = \dfrac{\pi }{{64}}5{(l - x)^2}\]
\[f(x) = \dfrac{{5\pi }}{{64}}{\left( {\dfrac{{4x}}{x} + x - x} \right)^2}\]
\[f(x)\] \[ = \] \[\dfrac{{5\pi }}{4} \times {\left( {\dfrac{{4x}}{2}} \right)^2} = \dfrac{{5\pi }}{4} \times \dfrac{{16{x^2}}}{{{\pi ^2}}}\]
\[f(x) = \dfrac{{5{x^2}}}{{4\pi }}\]
\[\therefore f(x) = \dfrac{5}{{4\pi }} \times 2h = \dfrac{{5x}}{{2\pi }}\]
For minimum value; \[f'(x) = 0\]
\[\dfrac{5}{{2\pi }}x = 0\]
\[x = 0\]
If \[x = 0\] \[f'(0) = \dfrac{5}{{2\pi }} > 0\]

Note: We have to prove that Radius of Circle is Half the Side of the Square, for Least Sum of Avas.
So, they can be done taking \[r\] of the circle and side (S) of Square.
Since, \[r = \dfrac{x}{{2\pi }}\] (radius of circle)
And \[s = \dfrac{{l - x}}{4}\] (Side of a Square)
And, \[x = \dfrac{{\pi l}}{{4 + \pi }}\]
So, \[r = \dfrac{l}{{2(4 + \pi )}}\] ……. (i)
And, \[a = \dfrac{l}{{(4 + \pi )}}\] ……. (ii)
From (i) and (ii) get Radius of Circle is Half the Side of Square.