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प्रश्न
Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.
उत्तर
\[\text { Let the length of a side of the square and radius of the circle be x and r, respectively .} \]
\[\text { It is given that the sum of the perimeters of square and circle is constant .} \]
\[ \Rightarrow 4x + 2\pi r = K .............\left( \text { Where K is some constant } \right)\]
\[ \Rightarrow x = \frac{\left( K - 2\pi r \right)}{4} ........ \left( 1 \right)\]
\[\text { Now,} \]
\[A = x^2 + \pi r^2 \]
\[ \Rightarrow A = \frac{\left( K - 2\pi r \right)^2}{16} + \pi r^2 .............\left[ \text { From eq. } \left( 1 \right) \right]\]
\[ \Rightarrow \frac{dA}{dr} = \frac{\left( K - 2\pi r \right)^2}{16} + \pi r^2 \]
\[ \Rightarrow \frac{dA}{dr} = \frac{2\left( K - 2\pi r \right) - 2\pi}{16} + 2\pi r\]
\[ \Rightarrow \frac{dA}{dr} = \frac{\left( K - 2\pi r \right) - \pi}{4} + 2\pi r\]
\[ \Rightarrow \frac{\left( K - 2\pi r \right) - \pi}{4} + 2\pi r = 0\]
\[ \Rightarrow \frac{\left( K - 2\pi r \right)\pi}{4} = 2\pi r\]
\[ \Rightarrow K - 2\pi r = 8r ............. \left( 2 \right)\]
\[\frac{d^2 A}{d x^2} = \frac{\pi^2}{2} + 2\pi > 0\]
\[\text { So, the sum of the areas, A is least when }K - 2\pi r = 8r . \]
\[\text { From eqs }. \left( 1 \right) \text { and }\left( 2 \right), \text { we get}\]
\[x = \frac{\left( K - 2\pi r \right)}{4}\]
\[ \Rightarrow x = \frac{8r}{4}\]
\[ \Rightarrow x = 2r\]
\[ \therefore\text { Side of the square = Diameter of the circle }\]
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