SATHEE: Application of Derivatives 4 Question 13

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) $2 x = (π + 4) r$

(b) $(4 - π) x = π r$

(d) $2 x = r$

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Answer:

Correct Answer: 14. (c)

Solution:

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

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

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

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

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

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

$\Rightarrow π x + 4 x = 2 \Rightarrow x = \frac{2}{π + 4}$

Now, from Eq. (i), we get

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

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

Application of Derivatives 4 Question 13

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Application of Derivatives 4 Question 13

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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