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:
- A \(x = 2r\)
- B \(2x = r\)
- C \(2x = \left( {\pi + 4} \right)r\)
- D \(\left( {4 - \pi } \right)x = \pi r\)
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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