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

Toolbox:
• 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}}$$(0-2x)] \qquad=2\big[1-\large\frac{x}{\sqrt{4a^2-x^2}}\big]-----(2) Step 2: Differentiating with respect to x we get P''(x)=2\bigg[\large\frac{0-\sqrt{4a^2-x^2}(1)-x.\large\frac{1}{2}(4a^2-x^2)^{\Large\frac{-1}{2}}(-2x)}{4(a^2-x^2)}\bigg] \qquad=2\bigg[\large\frac{-4a^2}{(4a^2-x^2)^{\Large\frac{3}{2}}}\bigg] \qquad=\bigg[\large\frac{-8a^2}{(4a^2-x^2)^{\Large\frac{3}{2}}}\bigg]-----(3) Step 3: For P(x) to be minimum P'(x)=0 and P''(x)<0 \Rightarrow 2\bigg[1-\large\frac{x}{\sqrt{4a^2-x^2}}\bigg]$$=0$
$\Rightarrow 1-\large\frac{x}{\sqrt{4a^2-x^2}}$$=0$
$\Rightarrow 1=\large\frac{x}{\sqrt{4a^2-x^2}}$
$\Rightarrow 4a^2-x^2=x^2$
$2x^2=4a^2$
$x^2=\large\frac{4a^2}{2}$
$x^2=2a^2$
$x=\pm \sqrt 2 a$
Step 4:
$P''(x)=\large\frac{-8a^2}{(4a^2-2a^2)^{\Large\frac{3}{2}}}$
$\qquad=\large\frac{-8a^2}{(2a^2)^{\Large\frac{3}{2}}}$
$\Rightarrow P(x)$ is maximum at $x=\sqrt 2a$
Step 5:
$y^2=4a^2-x^2$
$\quad=4a^2-2a^2$
$\quad=2a^2$
$y=\sqrt 2a$
Thus $x=y$
$\therefore$ the rectangle is a square of side $\sqrt 2a$