# A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

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• Area of a circle $=\pi r^2$
• Area of a rectangle $=lb$
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} Step 1: Perimeter of the window when the width of window is x and 2r is the length. \Rightarrow 2x+2r+\large\frac{1}{2}$$\times 2\pi r$$=10[Given] 2x+2r+\pi r=10 2x+r(2+\pi)=10------(1) For admitting the maximum light through the opening the area of the window must be maximum. A=Sum of areas of rectangle and semi-circle. Step 2: Area of circle=\pi r^2 Area of rectangle= l\times b=2\times r\times x A=2rx+\large\frac{1}{2}\pi r^2 \;\;=r[10-(\pi+2)r]+\large\frac{1}{2}\pi r^2 \;\;=10r-(\large\frac{1}{2}$$\pi+2)r^2$
For maximum area $\large\frac{dA}{dr}$$=0 and \large\frac{d^2A}{dr^2} is -ve. \Rightarrow 10-(\pi+4)r=0 (\pi+4)r=10 r=\large\frac{10}{\pi+4} Step 3: \large\frac{d^2A}{dr^2}$$=-(\pi+4)$[Differentiating with respect to r]
(i.e)$\large\frac{d^2A}{dr^2}$ is -ve for $r=\large\frac{10}{\pi+4}$
$\Rightarrow A$ is maximum.
From (1) we have
$\Rightarrow\:10=(\pi+2)r+2x$
Put the value of $r$ in (1)
$10=(\pi+2)\times \big(\large\frac{10}{\pi+4}\big)$$+2x 10=\large\frac{10(\pi+2)}{\pi+4}$$+2x$
$\;\;\;\;=\large\frac{10(\pi+2)+2x(\pi+4)}{\pi+4}$
$10(\pi+4)=10(\pi+2)+2x(\pi+4)$
$10(\pi+4)-10(\pi+2)=2x(\pi+4)$
$10\pi+40-10\pi-20=2x(\pi+4)$
$20=2x(\pi+4)$
$10=x(\pi+4)$
$x=\large\frac{10}{\pi+4}$
Step 4:
Length of rectangle$=2r=2\big(\large\frac{10}{\pi+4}\big)$
$\qquad\qquad\qquad\qquad\quad=\large\frac{20}{\pi+4}$
breadth=$\large\frac{10}{\pi+4}$
edited Feb 12, 2014