Discussion Forum

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

Answer:

Explanation:

According to the given information, we have

 the perimeter of square + perimeter of circle=2 units

  $\Rightarrow 4x+2\pi r=2\Rightarrow r=\frac{1-2x}{\pi}..(i)$

Now, let A be the sum of the areas of the square and the circle, Then

$A=x^{2}+\pi r^{2}$

   $A(x)=x^{2}+\pi\frac{(1-2x)^{2}}{\pi^{2}}$

 $\Rightarrow x^{2}+\frac{(1-2x)^{2}}{\pi^{}}$

Now, for minimum value of A(x). $\frac{\text{d}A}{\text{d}x}=0$

$\Rightarrow 2x+\frac{2(1-2x)}{\pi}.(-2)=0$

$\Rightarrow x=\frac{2-4x}{\pi}$

$\Rightarrow \pi x+4x=2$

$\Rightarrow x=\frac{2}{\pi+4}$   .........(ii)

Now, from Eq.(i), we get

$r=\frac{1-2.\frac{2}{\pi+4}}{\pi}$

$r=\frac{\pi+4-4}{\pi(\pi+4)}=\frac{1}{\pi+4}$     ..........(iii)

From Eq(ii) and (iii)

 we get x=2r