Question

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 ______.

Options

• `22/(9 + 4sqrt(3))`

• `66/(9 + 4sqrt(3))`

• `22/(4 + 9sqrt(3))`

• `66/(4 + 9sqrt(3))`

Answer 1

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 `underlinebb(66/(9 + 4sqrt(3))`.

Explanation:

Given: A wire of length 22 m Let length of the side of triangle be x and the length of the side of square be y and p be the length of wire formed into triangle

∴ p = 3x and 22 – p = 4y

⇒ x = `p/3` and y = `1/4(22 - p)`

Now, area of triangle = `sqrt(3)/4x^2 = sqrt(3)/4(p/3)^2`

and area of square = y² = `[1/4(22 - p)]^2`

∴ Total area = `sqrt(3)/4 p^2/9 + 1/16(22 - p)^2`

⇒ A = `sqrt(3)/36p^2 + 1/16(22^2 + p^2 - 44p)`

⇒ A = `sqrt(3)/36p^2 + p^2 - 22/8p + 22^2/16`

For A to be minimum, `(dA)/(dp)` = 0

⇒ `(dA)/(dp) = 2(sqrt(3)/36 p^2) + 2(p/16) - 22/8` = 0

⇒ `(dA)/(dp) = sqrt(3)/18p + p/8 - 22/8` = 0

⇒ `p(sqrt(3)/18 + 1/8) = 22/8`

⇒ p = `(22/8)/((4sqrt(3) + 9)/72)`

⇒ p = `22/8 xx 72/(4sqrt(3) + 9)`

⇒ x = `p/3 = (22 xx 9)/((4sqrt(3) + 9)3)`

⇒ x = `66/(4sqrt(3) + 9)`