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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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=$\large\frac{10}{\pi+4}$
answered Aug 12, 2013 by sreemathi.v
edited Feb 12, 2014 by rvidyagovindarajan_1
 

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