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) $2 x=r$
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Answer:
Correct Answer: 14. (d)
Solution:
- 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 $x=2 r$
