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.

Explanation

Let radius of semi-circle = r
`:. ` One side of rectangle = 2r and otherwise = x

`:.` P=perimeter = 10 (Given)
` rArr 2x + 2r + 1/2 (2pi r) = 10`
` rArr 2x = 10 - r (pi +2)` ….(1)
Again let A be the area of the window.
` :. ` A= area of semi - circle + area of rectangle = ` 1/2 pi r^(2) + 2 rx`
`rArr A = 1/2 pi r^(2) + r[10-r(pi+2)]` [ From eqn. (1)]
` 1/2 pi r^(2) + 10r-r^(2)pi - 2 r^(2)`
` = 10r - (pi r^(2))/2 - 2r^(2)`
` rArr (dA)/(dr)= 10- pi r - 4r` .....(2)
and ` (d^(2)A)/(dr^(2) = - pi - 4` ...(3)
For maxima/minima ` (dA)/(dr) = 0`
` rArr 10- pi r - 4r= 0 rArr 10 = (4+pi) r `
` rArr r = 10/(4+pi)`
put ` r = 10/(4+pi) " in equation (3) ",(d^(2)A)/(dr^(2)) lt 0 `
`:." A has local maximum value at " r = 10/(4+pi)`
`:. " radius of semicircle " r = 10/(4+pi)` ....(4)
and a side of rectangle = ` 2r = (2 xx 10)/(4+pi) = 20 /(4+ pi)`
From equation (1) other side of rectangle ` x = 1/2 [10 - r (pi + 2)]`
` 1/2 [10-(10/(pi+4))(pi+2)]` [ From eqn. (4)
` = (10pi+40-10pi-20)/(2(pi+4))`
` = 20/(2(pi+4)) = 10/(pi+4)`
`:.` The light emitting from the window will be maximum when the area will be maximum, so the dimensions of the window are
radius ` = r = 10/(pi+4),"length "=2r=20/(pi+4),"breadth "=x=10/(pi+4)`.