Application of Derivatives 4 Question 13

####14. 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

(2016 Main)

(a) $2 x=(\pi+4) r$

(b) $(4-\pi) x=\pi r$

(c) $x=2 r$

(d) $2 x=r$

Show Answer

Answer:

Correct Answer: 14. (c)

Solution:

  1. According to given information, we have Perimeter of square + Perimeter of circle $=2$ units

$ \begin{aligned} \Rightarrow & 4 x+2 \pi r & =2 \\ \Rightarrow & r & =\frac{1-2 x}{\pi} \end{aligned} $

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

$ \begin{aligned} A & =x^{2}+\pi r^{2} \\ & =x^{2}+\pi \frac{(1-2 x)^{2}}{\pi^{2}} \\ \Rightarrow \quad A(x) & =x^{2}+\frac{(1-2 x)^{2}}{\pi} \end{aligned} $

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

$\Rightarrow 2 x+\frac{2(1-2 x)}{\pi} \cdot(-2)=0 \Rightarrow x=\frac{2-4 x}{\pi}$

$ \Rightarrow \quad \pi x+4 x=2 \Rightarrow x=\frac{2}{\pi+4} $

Now, from Eq. (i), we get

$ r=\frac{1-2 \cdot \frac{2}{\pi+4}}{\pi}=\frac{\pi+4-4}{\pi(\pi+4)}=\frac{1}{\pi+4} $

From Eqs. (ii) and (iii), we get $x=2 r$



NCERT Chapter Video Solution

Dual Pane