Question

Show that all rectangles inscribed in a fixed circle square has maximum area.

Answer 1

Step $1$

Let length and breath of rectangle inscribed in a circle of radius $abc$ $x$ and $y$ respectively.

$x^2+y^2={2a}^2$

$x^2+y^2={4a}^2$

Perimeter $=2(x+y)$

$P\left(x\right)=x|x+\sqrt{{4a}^2-x^2}|$

$P^1\left(x\right)=z[1+\frac{1}{2\sqrt{{4a}^2-x^2}}.(-2x)]$

$=z[1-\frac{x}{\sqrt{{4a}^2-x^2}}]⟶\left(2\right)$

On double differentiation

$P^{11}\left(x\right)=2\left[\frac{0-\sqrt{{4a}^2-x^2}\left(1\right)-x.\frac{1}{2}{({4a}^2-x^2)}^{-1/2}(-2x)}{4(a^2-x^2)}\right]$

$=\left[\frac{{-8a}^2}{{({4a}^2-x^2)}^{3/2}}\right]⟶\left(3\right)$

For $P\left(x\right)$ to be minimum $P^1\left(x\right)=0$

$2[1-\frac{x}{\sqrt{{4a}^2-x^2}}]=0$

$1-\frac{x}{\sqrt{{4a}^2-x^2}}=0$

$x=\pm 2a$

From eq$\left(3\right)$ $P^{11}\left(x\right)=\frac{{-8a}^2}{{({4a}^2-{2a}^3)}^{3/2}}=\frac{{-8a}^2}{{\left({2a}^2\right)}^{3/2}}$

$P\left(x\right)$ maximum at $x=\sqrt{2}a$

From $\left(1\right)$ $y^2={4a}^2-x^2={4a}^2-{2a}^2={2a}^2=\sqrt{2}a$

Thus $x=y$ rectangle is a square of side $\sqrt{2}a$.