Question

Solve the following :  A window is in the form of a rectangle surmounted by a semicircle. If the perimeter be 30 m, find the dimensions so that the greatest possible amount of light may be admitted.

Answer 1

Let x be the length, y be the breadth of the rectangle and r be the radius of the semicircle. Then perimeter of the window

Then perimeter of the window = x + 2y + πr, where x = 2r

This is given to be m
∴ 2r + 2y + πr = 30
2y = 30 – (π + 2)r

∴ y = `15 - ((pi + 2)r)/(2)`           ...(1)
The greatest possible amount of light may be admitted if the area of the window is maximum. Let A be the area of the window.

Then A = `xy + (pir^2)/(2)`

= `2yr + (pir^2)/(2)`                        ...[∵ x = 2r]

= `2r[15 - ((pi + 2))/2] + (pir^2)/(2)`   ...[By (1)]

= `30r - (pi + 2)r^2 + pi/(2)r^2`

= `30r - (pi + 2 - pi/2)r^2`

∴ A = `30r - ((pi + 4)/2)r^2`

∴ `"dA"/"dr" = d/"dr"[30r - ((pi + 4)/2)r^2]`

= `30 xx 1 - ((pi + 4)/2) xx 2r`

= 30 – (π + 4)r

and
`(d^2"A")/"dr" = d/"dr"[30 -(pi + 4)r]`

= `0 - (pi + 4) xx 1`
= – (π + 4)

For maximum A, `"dA"/"dr"` = 0

∴ 30 – (π + 4)r = 0

∴ r = `(30)/(pi + 4)`
and
`((d^2"A")/("dr"))_("at"  r = (30)/(pi + 4)) = - (pi + 4) < 0`

∴ A is s maximum when r = `(30)/(pi + 4)`

When r = `(30)/(pi + 4) x = 2 = (60)/(pi + 4)`
and
y = `15 - ((pi + 2))/(2) xx (30)/(pi + 4)`        ...[By (1)]

= `(30pi + 120 - 30pi - 60)/(2(pi + 4)`

= `(30)/(pi + 4)`

Hence, the required dimensions of the window are as follows :

Length of rectangle = `((60)/(pi + 4))` metres,

breadth of rectangle = `((30)/(pi + 4))` metres and

radius of the semicircle = `((30)/(pi + 4))` metres.