Question

Show that the rectangle of maximum area that can be inscribed in a circle of radius '$r$' is a square of side $\sqrt{2r}$.

Answer 1

Let $ABCD$ be the rectangle inscribed in a circle of radius $r$

$AC$ and $BD$ are diameters of length $2r$ as angle is semicircle is always ${90}^o$

let $x$ be the length and $y$ be the breadth

$x^2+y^2=(2r{)}^2$

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

$y=\sqrt{4r^2-x^2}$

Area of rectangle $=xy$

$A=x\sqrt{4r^2-x^2}$

$\frac{dA}{dx}=\sqrt{4r^2-x^2}-\frac{2r^2}{2\sqrt{4r^2-x^2}}$

$\frac{dA}{dx}=\frac{4r^2-x^2-x^2}{\sqrt{4r^2-x^2}}=\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}$

for minimum area

$\frac{dA}{dx}=0$

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

$\Rightarrow 4r^2-2x^2=0$

$x=\sqrt{2}r$

also $x=\sqrt{2}r$

Since length and breadth are same,

So required rectangle of length and breadth is a square of $\sqrt{2}r$