Question

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

Answer 1

Let the radius of semicircle, length and breadth of rectangle be r, x and y respectively 

AE = r 

AB = x = 2r (semicircle is mounted over rectangle) …(1) 

AD = y 

Given : 

Perimeter of window = 10 m 

x + 2y + πr = 10 

⇒ 2r + 2y + πr = 10 

⇒ 2y = 10 – (π + 2).r

⇒ y = \(\frac{10-(\pi + 2)r}{2}\) ...(2)

To admit maximum amount of light, area of window should be maximum.

Assuming area of window as A,

A = xy + \(\frac{nr^2}{2}\)

⇒ A = (2r) (\(\frac{10-(\pi + 2)r}{2}\)) +\(\frac{nr^2}{2}\) 

⇒ A = 10r - πr² – 2r² + \(\frac{nr^2}{2}\)

⇒ A = 10r – 2r² - \(\frac{nr^2}{2}\)

Condition for maxima and minima is,

\(\frac{dA}{dr}\) = 0

⇒ 10 – 4r - πr = 0

⇒ r = \(\frac{10}{4+\pi}\) A will be maximum.

Length of rectangular part = \(\frac{20}{4+\pi}\)m (from equation 1)

Breath of rectangular part = \(\frac{10-(\pi + 2)r}{2}\) (from equation 2)