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The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

10) The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

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It is given that
the sum of the perimeter of a circle and square is k = 2\pi r + 4a = k\Rightarrow a = \frac{k - 2\pi r}{4} 
Let the sum of the area of a circle and square(A) = \pi r^2 + a^2
A = \pi r^2 + (\frac{k-2\pi r}{4})^2
A^{'}(r) = 2\pi r + 2(\frac{k-2\pi r}{16})(- 2\pi)\\ A^{'}(r) = 0\\ 2\pi (\frac{8r-k-2\pi r}{8}) = 0\\ r = \frac{k}{8-2\pi}
Now,
A^{''}(r) = 2\pi (\frac{8-2\pi }{8}) = 0\\ A^{''}(\frac{k}{8-2\pi}) > 0
Hence, r= \frac{k}{8-2\pi} is the point of minima
a = \frac{k-2\pi r}{4} = \frac{k-2\pi \frac{k}{8-2\pi}}{4}=2 \frac{k}{8-2\pi} = 2r
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

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