A window is in the shape of a rectangle surmounted by semi-circle. If the perimeter of the window is $20 f t$ , find the maximum area.

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Solution

Clearly, $r = \frac{l}{2}$
Perimeter =P(rectangle )+ P(semi-circle)
$= 2 (l + b) + π r$
$= r + 2 b + π r = 20$
$\Rightarrow b = \frac{20 - r (1 + π)}{2}$
Area, $A = l b + π r^{2}$
$= 2 r b + π r^{2}$
$= 2 r (20 - r - r π) + π r^{2} = 20 r - r^{2}$
$\Rightarrow \frac{d A}{d r} = 20 - 2 r = 0$ at $r = 10$ $\frac{d^{2} A}{d r^{2}} = - 2 < 0 \Rightarrow$ Maxima
$\Rightarrow A_{m a x} = 200 - 100 = 100 f t^{2}$

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A window is in the shape of a rectangle surmounted by semi-circle. If the perimeter of the window is 20 ft, find the maximum area.

Clearly, r=l2Perimeter =P(rectangle )+ P(semi-circle)=2(l+b)+πr=r+2b+πr=20⇒b=20−r(1+π)2Area, A=lb+πr2 =2rb+πr2=2r(20−r−rπ)+πr2=20r−r2⇒dAdr=20−2r=0 at r=10 d2Adr2=−2<0⇒ Maxima⇒Amax=200−100=100 ft2

A window is in the shape of a rectangle surmounted by semi-circle. If the perimeter of the window is $20 f t$ , find the maximum area.