Question

A window is in the form of rectangle above which there is a semi circle. If the perimeter of window is p units, show that the window will allow maximum light, when radius of the semicircle is \( \large\frac{p}{\pi + 4} \) units.

Answer 1

hence this is proved........

Answer 2

r = p/(4 +  π)

Step-by-step explanation:

Let say radius = r  so one side  = 2r

Another side = a

Perimeter of window =  2r + 2a  +  πr = p

p = 2r + 2a  +  πr

p = r(2 + π) + 2a

=> a = (p -  r(2 + π))/2

Maximum light when area is maximum

Area = πr²/2  + 2ar

A =  πr²/2  + 2 r(p -  r(2 + π))/2

A = πr²/2 +  r(p -  r(2 + π))

A = πr²/2 + rp - r²(2 + π)

dA/dr = πr + p - 2r(2 + π)

dA/dr = πr + p - 4r  - 2 πr

πr + p - 4r  - 2 πr = 0

=> p = r(4 + π)

=> r = p/(4 +  π)

d²A/dr² = π  -4  - 2 π  (-ve)

Hence maximum area

Learn more:

this problem is about maxima and minima. find the area of the ...

https://brainly.in/question/13355753