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प्रश्न
A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.
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उत्तर
\[\text { Let the dimensions of the rectangle be x and y } . \]
\[\text { Perimeter of the window = x + y + x + x + y = 12 }\]
\[ \Rightarrow 3x + 2y = 12\]
\[ \Rightarrow y = \frac{12 - 3x}{2} ...........\left( 1 \right)\]
\[\text { Area of the window } =xy+\frac{\sqrt{3}}{4} x^2 \]
\[ \Rightarrow A = x\left( \frac{12 - 3x}{2} \right) + \frac{\sqrt{3}}{4} x^2 \]
\[ \Rightarrow A = 6x - \frac{3 x^2}{2} + \frac{\sqrt{3}}{4} x^2 \]
\[ \Rightarrow \frac{dA}{dx} = 6 - \frac{6x}{2} + \frac{2\sqrt{3}}{4}x\]
\[ \Rightarrow \frac{dA}{dx} = 6 - 3x + \frac{\sqrt{3}}{2}x\]
\[ \Rightarrow \frac{dA}{dx} = 6 - x\left( 3 - \frac{\sqrt{3}}{2} \right)\]
\[\text { For maximum or a minimum values of A, we must have }\]
\[\frac{dA}{dx} = 0\]
\[ \Rightarrow 6 = x\left( 3 - \frac{\sqrt{3}}{2} \right)\]
\[ \Rightarrow x = \frac{12}{6 - \sqrt{3}}\]
\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]
\[y = \frac{12 - 3\left( \frac{12}{6 - \sqrt{3}} \right)}{2}\]
\[ \Rightarrow y = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}\]
\[\text { Now, }\]
\[\frac{d^2 A}{d x^2} = - 3 + \frac{\sqrt{3}}{2} < 0\]
\[\text { Thus, the area is maximum when }x=\frac{12}{6 - \sqrt{3}}\text { and }y=\frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}.\]
Notes
The solution given in the book is incorrect. The solution here is created according to the question given in the book.
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