JEE Mains · Maths · STD 12 - 6. Application of derivatives
A wire of length \(20\, \mathrm{~m}\) is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in \(meters\)) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:
- A \(\frac{5}{2+\sqrt{3}}\)
- B \(\frac{10}{2+3 \sqrt{3}}\)
- C \(\frac{5}{3+\sqrt{3}}\)
- D \(\frac{10}{3+2 \sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{10}{2+3 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Let the wire is cut into two pieces of length \(\mathrm{x}\) and \(20-x\). Area of square \(=\left(\frac{\mathrm{x}}{4}\right)^{2}\) Area of regular hexagon \(=6 \times \frac{\sqrt{3}}{4}\left(\frac{20-x}{6}\right)^{2}\) Total area…
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