Explanation

(d)
Area of rectangle, $A = 2x.2\sqrt{r^{2} - x^{2}}$
$= 4x\sqrt{r^{2} - x^{2}}$
$\frac{dA}{dx} = \frac{4(r^{2} - 2x^{2})}{\sqrt{r^{2} - x^{2}}}$
[image:6_image25.emf]
For maximum or minimum put $\frac{dA}{dx} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ x = \frac{r}{\sqrt{2}}$
It can be easily checked that $\frac{d^{2}A}{dx^{2}} < 0$ for $x = \frac{r}{\sqrt{2}}$
The maximum value of $A$ is given by
$A = 4\frac{r}{\sqrt{2}}\sqrt{r^{2} - \frac{r^{2}}{2}} = 2r^{2}$