The maximum area (in square units) of a rectangle whose vertices lie on the ellipse $x^{2} + y^{2} = 1$ is
1

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Solution

The correct option is A 1

Let $2 x, 2 y$ be the length and breadth respectively of the rectangle inscribed in the ellipse $x^{2} + 4 y^{2} = 1$
$\Rightarrow$ Area of rectangle $= (2 x) (2 y) = 4 x y$
Let $f = (A r e a)^{2} = 16 x^{2} y^{2} = 4 x^{2} (1 - x^{2})$
$f^{'} (x) = 8 x - 16 x^{3} = 0$
$\Rightarrow x = 0, \pm \frac{1}{\sqrt{2}}$
$f^{''} (x) = 8 - 48 x^{2}$
$f^{''} (\frac{1}{\sqrt{2}}) = 8 - 12 < 0$
$∴ f$ is maximum at $x = 1 / \sqrt{2}$
At $x = \frac{1}{\sqrt{2}}, y = \frac{1}{\sqrt{8}}$
Max. area $= 4 (\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{8}}) = 1$

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The maximum area (in square units) of a rectangle whose vertices lie on the ellipse x2+y2=1 is 1

The maximum area (in square units) of a rectangle whose vertices lie on the ellipse $x^{2} + y^{2} = 1$ is
1