Question

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius r is the square of side r√2.

Answer 1

Let ABCD be a rectangle inscribed in a circle of radius r with centre at O.

Let AB = 2x and BC = 2y be the sides of the rectangle.

Then in right angled ΔOAM, 

AM² + OM² = OA² ( By Pythagoras theorem)

⇒ x² + y² = r²

⇒ \(y=\sqrt{r^2-x^2}\) ...(i)

Let P be the perimeter of rectangle ABCD, then

P = 4x + 4y

⇒ \(P = 4x + 4 \sqrt{r^2-x^2}\)

⇒ \(\frac{dP}{dx}=\) \(4-\frac{4x}{\sqrt{r^2-x^2}}\)

For maximum or minimum value of P, we have 
\(\frac{dP}{dx}=\) 0

Thus, P is maximum when \(x=\frac{r}{\sqrt2}.\)

Putting \(x=\frac{r}{\sqrt2}\) in (i)_, we get

\(y=\frac{r}{\sqrt2}\)

Therefore,

x = y

⇒ 2x = 2y

Hence, P is maximum when rectangle is square of side \(2x = \sqrt{2}r.\)