JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Application of Derivatives

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A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :

(A) $\frac{22}{9 + 4 3}$
(B) $\frac{66}{9 + 4 3}$ ✓ Correct
(C) $\frac{22}{4 + 9 3}$
(D) $\frac{66}{4 + 9 3}$

Answer: (B)

Solution

Let the side of the equilateral triangle be

a

and the side of the square be

b

. Then

3 a = x

and

4 b = 22 - x

, where

x

is the length of wire used for the triangle.

The total area is

A_{T} = \frac{3}{4} a^{2} + b^{2} = \frac{3}{4} \cdot \frac{x ^{2}}{9} + \frac{( 22 - x ) ^{2}}{16}

.

For minimum area,

\frac{d A}{d x} = 0

\frac{3 x}{18} + \frac{2 ( 22 - x ) ( - 1 )}{16} = 0

x (\frac{3}{18} + \frac{1}{8}) = \frac{22}{8}

x \cdot \frac{4 3 + 9}{36} = \frac{11}{2}

x = \frac{11 \cdot 36}{2 ( 4 3 + 9 )} = \frac{198}{4 3 + 9}

Since

a = x /3

a = \frac{198}{3 ( 4 3 + 9 )} = \frac{66}{4 3 + 9} = \frac{66}{9 + 4 3}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Application of Derivatives

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