PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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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 - PUC Karnataka Science Class 12 - Mathematics

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Question

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.

Solution

\[\text { Let the dimensions of the rectangular part be x and y }.\]

\[\text { Radius of semi-circle } =\frac{x}{2}\]

\[\text { Total perimeter } = 10\]

\[ \Rightarrow \left( x + 2y \right) + \pi\left( \frac{x}{2} \right) = 10\]

\[ \Rightarrow 2y = \left[ 10 - x - \pi\left( \frac{x}{2} \right) \right]\]

\[ \Rightarrow y = \frac{1}{2}\left[ 10 - x\left( 1 + \frac{\pi}{2} \right) \right] ............ \left( 1 \right)\]

\[\text { Now }, \]

\[\text { Area }, A = \frac{\pi}{2} \left( \frac{x}{2} \right)^2 + xy\]

\[ \Rightarrow A = \frac{\pi x^2}{8} + \frac{x}{2}\left[ 10 - x\left( 1 + \frac{\pi}{2} \right) \right] .............\left[ \text { From eq. } \left( 1 \right) \right]\]

\[ \Rightarrow A = \frac{\pi x^2}{8} + \frac{10x}{2} - \frac{x^2}{2}\left( 1 + \frac{\pi}{2} \right)\]

\[ \Rightarrow \frac{dA}{dx} = \frac{\pi x}{4} + \frac{10}{2} - \frac{2x}{2}\left( 1 + \frac{\pi}{2} \right)\]

\[\text { For maximum or minimum values of A, we must have }\]

\[\frac{dA}{dx} = 0\]

\[ \Rightarrow \frac{\pi x}{4} + \frac{10}{2} - \frac{2x}{2}\left( 1 + \frac{\pi}{2} \right) = 0\]

\[ \Rightarrow x\left[ \frac{\pi}{4} - 1 - \frac{\pi}{2} \right] = - 5\]

\[ \Rightarrow x = \frac{- 5}{\left( \frac{- 4 - \pi}{4} \right)}\]

\[ \Rightarrow x = \frac{20}{\left( \pi + 4 \right)}\]

\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]

\[y = \frac{1}{2}\left[ 10 - \left( \frac{20}{\pi + 4} \right)\left( 1 + \frac{\pi}{2} \right) \right]\]

\[ \Rightarrow y = 5 - \frac{10\left( \pi + 2 \right)}{2\left( \pi + 4 \right)}\]

\[ \Rightarrow y = \frac{5\pi + 20 - 5\pi - 10}{\left( \pi + 4 \right)}\]

\[ \Rightarrow y = \frac{10}{\left( \pi + 4 \right)}\]

\[\frac{d^2 A}{d x^2} = \frac{\pi}{4} - \frac{\pi}{2} - 1\]

\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{\pi - 2\pi - 4}{4}\]

\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{- \pi - 4}{4} < 0\]

\[\text { Thus, the area is maximum when x= }\frac{20}{\pi + 4}\text { and } y=\frac{10}{\pi + 4}.\]

\[\text { So, the required dimensions are given below }: \]

\[\text { Length } = \frac{20}{\pi + 4} m\]

\[\text { Breadth }=\frac{10}{\pi + 4}m\]

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Solution 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 Concept: Graph of Maxima and Minima.
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