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JEE Mains · Maths · STD 12 - 6. Application of derivatives

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 ofthe square and the circle so formed is minimum, then: 

  1. A \(x = 2r\)
  2. B \(2x = r\)
  3. C \(2x = \left( {\pi + 4} \right)r\)
  4. D \(\left( {4 - \pi } \right)x = \pi r\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x = 2r\)

Step-by-step Solution

Detailed explanation

given that \(4x + 2\pi r = 2\) i.e. \(2x + \pi r = 1\) \(\therefore r = \frac{{1 - 2x}}{\pi }\,\,\,\,\,\,.....\left( i \right)\) Area \(A = {x^2} + \pi {r^2}\) \( = {x^2} + \frac{1}{\pi }{\left( {2x - 1} \right)^2}\) for min value of area \(A\) \(\frac{{dA}}{{dx}} = 0\) given…
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