Let ABCD be the rectangle which is inscribed in a semi - circle with centre O and radius r.

Assume that the length of rectangle is 2x and breadth is 2y.

⇒ CD = AB = 2x

**In right angle ΔACD,We get **

x^{2 }+ 4y^{2 }= r^{2 }..**.(i)**

**(By Pythagoras theorem)**

**Now area of rectangle ABCD is given by,**

**A = length x breadth**

⇒ \(A=2x\times2y=4xy\)

Let Z = A^{2} = 4(r^{2}x^{2} - x^{4}) ..**.(ii)**

Then Z is maximum or minimum accordings as A is maximum or minimum.

**Differentiating equation (ii) with respect to x, we get**

**\(\frac{dZ}{dx}=\) **4[r^{2} . 2x - 4x^{3}]

**For maximum or minimum value of Z, we have**

**\(\frac{dZ}{dx}=0\) **

⇒ r^{2} - 2x^{2} = 0

**( ∵ x cannot be zero)**

⇒ \(x^2=\frac{r^2}{2}\)

**Now,**

**Thus area will be maximum when,**

**\(x=\frac{r}{\sqrt 2}\)**

⇒ \(2x =\sqrt 2r,\)

**Putting \(x=\frac{r}{\sqrt 2}\) in equation (i), we obtain,**

**\(y=\frac{r}{2\sqrt 2}\) **

⇒ 2y = **\(\frac{r}{\sqrt 2}\) **

**Now,**

A = \(2x \times 2y\)

=** \(\sqrt2r\,\times\,\frac{r}{\sqrt2}\) **

= r^{2 }

**Hence, the dimensions are \(\sqrt2r\) and ****\(\frac{r}{\sqrt2}\)** and area = r^{2} sq units.