# Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

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• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} • Perimeter=2ab Step 1: Let the length and breadth of the rectangle inscribed in a circle of radius 'a' be 'x' and 'y' respectively. x^2+y^2=(2a)^2-----(1) x^2+y^2=4a^2 Perimeter=2(x+y) P(x)=2[x+\sqrt{4a^2-x^2}] Differentiating with respect to x we get P'(x)=2[1+\large\frac{1}{2}.\frac{1}{\sqrt{4a^2-x^2}}$$(0-2x)]$
$\qquad=2\big[1-\large\frac{x}{\sqrt{4a^2-x^2}}\big]$-----(2)
Step 2:
On double differentiation we get
$P''(x)=2\bigg[\large\frac{0-\sqrt{4a^2-x^2}(1)-x.\large\frac{1}{2}(4a^2-x^2)^{\Large\frac{-1}{2}}(-2x)}{4(a^2-x^2)}\big]$
$\qquad=2\bigg[\large\frac{-4a^2}{(4a^2-x^2)^{\Large\frac{3}{2}}}\bigg]$
$\qquad=\bigg[\large\frac{-8a^2}{(4a^2-x^2)^{\Large\frac{3}{2}}}\bigg]$-----(3)
Step 3:
For $P(x)$ to be minimum $P'(x)=0$ and $P''(x)<0$
From (1),$P'(x)=0$
$\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$
$\Rightarrow 1=\large\frac{x}{\sqrt{4a^2-x^2}}$
$\Rightarrow 4a^2-x^2=x^2$
$2x^2=4a^2$
$x^2=\large\frac{4a^2}{2}$
$x^2=2a^2$
$x=\pm \sqrt 2 a$
Taking only positive value $x=\sqrt 2a$
Step 4:
From equation (3)
$P''(x)=\large\frac{-8a^2}{(4a^2-2a^2)^{\Large\frac{3}{2}}}$
$\qquad=\large\frac{-8a^2}{(2a^2)^{\Large\frac{3}{2}}}$ which is +ve
$\Rightarrow P(x)$ is maximum at $x=\sqrt 2a$
Step 5:
From (1)
$y^2=4a^2-x^2$
$\quad=4a^2-2a^2$
$\quad=2a^2$
$y=\sqrt 2a$
Thus $x=y$
Hence the rectangle is a square of side $\sqrt 2a$
edited Aug 19, 2013