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

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.