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(A) Let the radius of the semi-circle be $R = 2$. Let the rectangle have length $2x$ and breadth $y$ inscribed in the semi-circle.By the Pythagorean t

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$\sqrt{2}$

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(A) Let the radius of the semi-circle be $R = 2$. Let the rectangle have length $2x$ and breadth $y$ inscribed in the semi-circle.By the Pythagorean theorem, $x^2 + y^2 = R^2 = 2^2 = 4$.Thus, $x = \sqrt{4 - y^2}$.The area of the rectangle is $A = (2x) 	imes y = 2y \sqrt{4 - y^2}$.To maximize $A$, we maximize $A^2 = 4y^2(4 - y^2) = 16y^2 - 4y^4$.Let $f(y) = 16y^2 - 4y^4$. Differentiating with respect to $y$, we get $f'(y) = 32y - 16y^3$.Setting $f'(y) = 0$, we have $16y(2 - y^2) = 0$.Since $y > 0$, $y^2 = 2$, so $y = \sqrt{2}$.Then $x = \sqrt{4 - 2} = \sqrt{2}$.The sides of the rectangle are $2x = 2\sqrt{2}$ and $y = \sqrt{2}$.The smaller side is $\sqrt{2}$.

The smaller side of the rectangle with the largest area, that can be inscribed inside a semi-circle of radius $2 \ units$ is of length

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