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A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the - Mathematics

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प्रश्न

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

योग

उत्तर

Let the length of one piece be x m and other piece is of length (28 - x) m Let the length of the piece bent into the shape of a circle be x m and length of the other piece bent into the shape of a square is (28 - x) m.

Circumference = 2πr

⇒ 2πr = x

⇒ `r = x/(2pi)`

Area of the circle= π (radius)2

`= pi (x/(2pi))^2 = x^2/(4pi)`

Perimeter of square = 4 side

⇒ 28 - x = 4 side

⇒ side = `(28 - x)/4`

⇒ Area of the square = (side)2

`= ((28 - x)/4)^2`

`= (28 - x)^2/16`

Let A be the sum of the areas of the two figures, then

`A = x^2/(4pi) + (28 - x)^2/16`

Differentiating w.r.t. x, we get

`(dA)/dx = (2x)/(4pi) + (2 (28 - x)(-1))/16`

`= x/(2pi) - (28 - x)/8`

For maximum / minimum, `(dA)/dx = 0`

⇒ `x / (2pi) - (28 - x)/8 = 0`

⇒ ` (4x - 28pi + xpi)/(8pi) = 0`

⇒ `4x + xpi = 28 pi`

⇒ `x = (28pi)/ (4 + pi)`

⇒ `(d^2A)/dx^2 = 1/(2pi) - (-1)/8 = 1/ (2pi) + 1/8`

and `((d^2A)/dx^2)_(x = (28pi)/(4+pi))`

`= 1/(2pi) + 1/8 > 0`

Hence area A is minimum

∴ The wire must be cut at a distance of `(28pi)/(4+pi)` m. from one end.

Hence, the length of the two pieces are `(28pi)/(4 + pi)` m and `(28 - (28pi)/(4+pi)) m  112/(4 + pi)`  m

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अध्याय 6: Application of Derivatives - Exercise 6.5 [पृष्ठ २३३]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.5 | Q 22 | पृष्ठ २३३

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