Question

A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is $20$ ft., The maximum area is $\frac{200}{\pi +k}$. Find k

Answer 1

perimeter of window $=20ft$

perimeter $=l+2b+\pi \frac{l}{2}=20$

$(1+\frac{\pi }{2})l+2b=20$...........(1)

$A=l.b+\frac{\pi }{2}{\left(\frac{l}{2}\right)}^2=\frac{\pi }{4}{l}^2+lb$

for area to be max.

$dA=\frac{\pi }{2}l\frac{dl}{dA}+b\frac{dl}{dA}+l\frac{dl}{dA}=0$

from equation (1), we have $\frac{dl}{dA}=\frac{-1}{2}(1+\frac{\pi }{2})\frac{dl}{dA}$

So, $[\frac{\pi }{2}l+b-\frac{1}{2}(1+\frac{\pi }{2})l]\frac{dl}{dA}=0\Rightarrow \frac{b}{l}=\frac{1}{2}(1-\frac{\pi }{2})$

solving this gives $l=10$

anf further solving gives $k=4$.