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# Solution for 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. - CBSE (Science) Class 12 - Mathematics

#### Question

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.

#### Solution

$\text { Let the length of a side of the square and radius of the circle be x and r, respectively .}$

$\text { It is given that the sum of the perimeters of square and circle is constant .}$

$\Rightarrow 4x + 2\pi r = K \left( \text { Where K is some constant } \right)$

$\Rightarrow x = \frac{\left( K - 2\pi r \right)}{4} . . . \left( 1 \right)$

$\text { Now,}$

$A = x^2 + \pi r^2$

$\Rightarrow A = \frac{\left( K - 2\pi r \right)^2}{16} + \pi r^2 \left[ \text { From eq } . \left( 1 \right) \right]$

$\Rightarrow \frac{dA}{dr} = \frac{\left( K - 2\pi r \right)^2}{16} + \pi r^2$

$\Rightarrow \frac{dA}{dr} = \frac{2\left( K - 2\pi r \right) - 2\pi}{16} + 2\pi r$

$\Rightarrow \frac{dA}{dr} = \frac{\left( K - 2\pi r \right) - \pi}{4} + 2\pi r$

$\Rightarrow \frac{\left( K - 2\pi r \right) - \pi}{4} + 2\pi r = 0$

$\Rightarrow \frac{\left( K - 2\pi r \right)\pi}{4} = 2\pi r$

$\Rightarrow K - 2\pi r = 8r . . . \left( 2 \right)$

$\frac{d^2 A}{d x^2} = \frac{\pi^2}{2} + 2\pi > 0$

$\text { So, the sum of the areas, A is least when }K - 2\pi r = 8r .$

$\text { From eqs }. \left( 1 \right) \text { and }\left( 2 \right), \text { we get}$

$x = \frac{\left( K - 2\pi r \right)}{4}$

$\Rightarrow x = \frac{8r}{4}$

$\Rightarrow x = 2r$

$\therefore\text { Side of the square = Diameter of the circle }$

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Solution for question: 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. concept: Graph of Maxima and Minima. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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