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
x2 + 4y2 = r2 ...(i)
(By Pythagoras theorem)
Now area of rectangle ABCD is given by,
A = length x breadth
⇒ \(A=2x\times2y=4xy\)
Let Z = A2 = 4(r2x2 - x4) ...(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[r2 . 2x - 4x3]
For maximum or minimum value of Z, we have
\(\frac{dZ}{dx}=0\)
⇒ r2 - 2x2 = 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 = \(2x \times 2y\)
= \(\sqrt2r\,\times\,\frac{r}{\sqrt2}\)
= r2
Hence, the dimensions are \(\sqrt2r\) and \(\frac{r}{\sqrt2}\) and area = r2 sq units.