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A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light - Mathematics

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प्रश्न

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening

योग

उत्तर

Let x and y be the length and breadth of the rectangle.

Radius of the semi - circle `= x/2`

Circumference of the semi - circle = `(pix)/2.`

Perimeter of the window

AB + BC + AD + DC

`x + 2y + (pix)/2= 10`

⇒ 2x + 4y + πx = 20

⇒ `y = (20 - (2 + pi)x)/4`

Area of the window = area of rectangle + area of a semicircle.

`A = xy + 1/2 pi (x/2)^2`

`= x ((20 - (2 + pi)x)/4) + (pix^2)/8.`

`A = (20x - (2 + pi) x^2)/4 + (pix^2)/8.`

∴ `(dA)/dx = (20 - (2 + pi) 2x)/4 + (2pix)/8`

For maxima / minima of A, 

`(dA)/dx = 0`

⇒ `(20 - (2 + pi) 2x)/4 + (2pix)/8 = 0`

⇒ 20 - (2 + π) 2x + πx = 0

⇒ 20 + x (π - 4 - 2π) = 0

⇒ 20 - x (4 + π) = 0

⇒ `x = 20/ (4 + pi)`

`(d^2A)/dx^2 = (-(2 + pi)2)/4 + (2pi)/8`

`= (-4 -2pi + pi)/4`

` = (-4 -pi)/4`

⇒ `(d^2A)/dx^2 < 0`

Hence the window admit the maximum light when x = length =  `20/ (4 + pi)`

and breadth `y = (20 - (2 + pi) 20/(4 + pi))/4`

`= (80 + 20pi - 40 - 20 pi)/(4 (4 + pi))`

`= 40/ (4(4 + pi))`

`= 10/ (4 + pi).`

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अध्याय 6: Application of Derivatives - Exercise 6.6 [पृष्ठ २४३]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.6 | Q 11 | पृष्ठ २४३

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