ExamBro
ExamBro
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:

  1. A \(\frac{5}{2+\sqrt{3}}\)
  2. B \(\frac{10}{2+3 \sqrt{3}}\)
  3. C \(\frac{5}{3+\sqrt{3}}\)
  4. D \(\frac{10}{3+2 \sqrt{3}}\)
Verified Solution

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…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app