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Question
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
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Solution
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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