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Show that the rectangle of maximum area that can be inscribed in a circle is a square.

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  • Perimeter=$2ab$
Step 1:
Let the length and breadth of the rectangle inscribed in a circle of radius 'a' be 'x' and 'y' respectively.
Differentiating with respect to x we get
Step 2:
Differentiating with respect to x we get
Step 3:
For $P(x)$ to be minimum $P'(x)=0$ and $P''(x)<0$
$\Rightarrow 2\bigg[1-\large\frac{x}{\sqrt{4a^2-x^2}}\bigg]$$=0$
$\Rightarrow 1-\large\frac{x}{\sqrt{4a^2-x^2}}$$=0$
$\Rightarrow 1=\large\frac{x}{\sqrt{4a^2-x^2}}$
$\Rightarrow 4a^2-x^2=x^2$
$x=\pm \sqrt 2 a$
Step 4:
$\Rightarrow P(x)$ is maximum at $x=\sqrt 2a$
Step 5:
$y=\sqrt 2a$
Thus $x=y$
$\therefore$ the rectangle is a square of side $\sqrt 2a$
answered Sep 27, 2013 by sreemathi.v
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