Q. 64.5( 2 Votes )

A window in the f

Answer :


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 = … 2


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


Assuming area of window as A


A = xy +


A = (2r) () +


A = 10r - πr2 – 2r2 +


A = 10r – 2r2 -


Condition for maxima and minima is



10 – 4r - πr = 0



= - 4 - π < 0


For r = A will be maximum.


Length of rectangular part = (from equation 1)


Breath of rectangular part = (from equation 2)



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