JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Application of Derivatives

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A wire of length 20 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) $\frac{5}{3 + 3}$
(B) $\frac{10}{3 + 2 3}$ ✓ Correct
(C) $\frac{10}{2 + 3 3}$
(D) $\frac{5}{2 + 3}$

Answer: (B)

Solution

Let the square use

x

m and the hexagon

20 - x

m of wire, so the square has side

\frac{x}{4}

and the hexagon side

\frac{20 - x}{6}

. The total area is

A = \frac{x ^{2}}{16} + 6 \cdot \frac{3}{4} (\frac{20 - x}{6})^{2} = \frac{x ^{2}}{16} + \frac{3}{24} (20 - x)^{2} .

Then

\frac{d A}{d x} = \frac{x}{8} - \frac{3}{12} (20 - x) = 0

gives

x (3 + 23) = 403

, so

20 - x = \frac{60}{3 + 2 3}

. The side of the hexagon is

\frac{20 - x}{6} = \frac{10}{3 + 2 3}

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Application of Derivatives

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