JEE Mains · Maths · STD 12 - 6. Application of derivatives
A wire of length \(36\, \mathrm{~m}\) is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is \(\mathrm{k}\) \((meter),\) then \(\left(\frac{4}{\pi}+1\right) \mathrm{k}\) is equal to ..... .
- A \(306\)
- B \(36\)
- C \(144\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(36\)
Step-by-step Solution
Detailed explanation
Let \(x+y=36\) \(\mathrm{x}\) is perimeter of square and \(\mathrm{y}\) is perimeter of circle side of square \(=x / 4\) radius of circle \(=\frac{y}{2 \pi}\) Sum Areas \(=\left(\frac{x}{4}\right)^{2}+\pi\left(\frac{y}{2 \pi}\right)^{2}\)…
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