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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?
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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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