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प्रश्न
A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.
उत्तर
\[\text { Let the dimensions of the rectangular part be x and y }.\]
\[\text { Radius of semi-circle } =\frac{x}{2}\]
\[\text { Total perimeter } = 10\]
\[ \Rightarrow \left( x + 2y \right) + \pi\left( \frac{x}{2} \right) = 10\]
\[ \Rightarrow 2y = \left[ 10 - x - \pi\left( \frac{x}{2} \right) \right]\]
\[ \Rightarrow y = \frac{1}{2}\left[ 10 - x\left( 1 + \frac{\pi}{2} \right) \right] ............ \left( 1 \right)\]
\[\text { Now }, \]
\[\text { Area }, A = \frac{\pi}{2} \left( \frac{x}{2} \right)^2 + xy\]
\[ \Rightarrow A = \frac{\pi x^2}{8} + \frac{x}{2}\left[ 10 - x\left( 1 + \frac{\pi}{2} \right) \right] .............\left[ \text { From eq. } \left( 1 \right) \right]\]
\[ \Rightarrow A = \frac{\pi x^2}{8} + \frac{10x}{2} - \frac{x^2}{2}\left( 1 + \frac{\pi}{2} \right)\]
\[ \Rightarrow \frac{dA}{dx} = \frac{\pi x}{4} + \frac{10}{2} - \frac{2x}{2}\left( 1 + \frac{\pi}{2} \right)\]
\[\text { For maximum or minimum values of A, we must have }\]
\[\frac{dA}{dx} = 0\]
\[ \Rightarrow \frac{\pi x}{4} + \frac{10}{2} - \frac{2x}{2}\left( 1 + \frac{\pi}{2} \right) = 0\]
\[ \Rightarrow x\left[ \frac{\pi}{4} - 1 - \frac{\pi}{2} \right] = - 5\]
\[ \Rightarrow x = \frac{- 5}{\left( \frac{- 4 - \pi}{4} \right)}\]
\[ \Rightarrow x = \frac{20}{\left( \pi + 4 \right)}\]
\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]
\[y = \frac{1}{2}\left[ 10 - \left( \frac{20}{\pi + 4} \right)\left( 1 + \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow y = 5 - \frac{10\left( \pi + 2 \right)}{2\left( \pi + 4 \right)}\]
\[ \Rightarrow y = \frac{5\pi + 20 - 5\pi - 10}{\left( \pi + 4 \right)}\]
\[ \Rightarrow y = \frac{10}{\left( \pi + 4 \right)}\]
\[\frac{d^2 A}{d x^2} = \frac{\pi}{4} - \frac{\pi}{2} - 1\]
\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{\pi - 2\pi - 4}{4}\]
\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{- \pi - 4}{4} < 0\]
\[\text { Thus, the area is maximum when x= }\frac{20}{\pi + 4}\text { and } y=\frac{10}{\pi + 4}.\]
\[\text { So, the required dimensions are given below }: \]
\[\text { Length } = \frac{20}{\pi + 4} m\]
\[\text { Breadth }=\frac{10}{\pi + 4}m\]
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