MHT CET · Maths · Application of Derivatives
A wire of length 8 units is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is
- A \(\frac{8}{\pi+4}\)
- B \(\frac{64}{\pi+4}\)
- C \(\frac{2}{\pi+4}\)
- D \(\frac{16}{\pi+4}\)
Answer & Solution
Correct Answer
(D) \(\frac{16}{\pi+4}\)
Step-by-step Solution
Detailed explanation
Let the length of the wire for the square be \(x\) and for the circle be \(8-x\). Area of square: \(A_s = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}\)
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