Show that among all positive numbers

Solution:

Here,

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

$\Rightarrow y=\sqrt{r^{2}-x^{2}}$               ......(1)

Now,

$Z=x+y$

$\Rightarrow Z=x+\sqrt{r^{2}-x^{2}}$                          [From eq. (1)]

$\Rightarrow \frac{d Z}{d x}=1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}$

For maximum or minimum values of $Z$, we must have

$\frac{d Z}{d x}=0$

$\Rightarrow 1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}=0$

$\Rightarrow 2 x=2 \sqrt{r^{2}-x^{2}}$

$\Rightarrow x=\sqrt{r^{2}-x^{2}}$

Squaring both the sides, we get

$\Rightarrow x^{2}=r^{2}-x^{2}$

$\Rightarrow 2 x^{2}=r^{2}$

$\Rightarrow x=\frac{r}{\sqrt{2}}$

Substituting the value of $x$ in eq. $(1)$, we get

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

$\Rightarrow y=\sqrt{r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}}$

$\Rightarrow y=\frac{r}{\sqrt{2}}$

$\frac{d^{2} z}{d x^{2}}=\frac{-\sqrt{r^{2}-x^{2}}+\frac{x(-x)}{\sqrt{r^{2}-x^{2}}}}{r^{2}-x^{2}}$

$\Rightarrow \frac{d^{2} z}{d x^{2}}=\frac{-r^{2}+x^{2}-x^{2}}{\left(r^{2}-x^{2}\right)^{\frac{3}{2}}}$

$\Rightarrow \frac{d^{2} z}{d x^{2}}=\frac{-r^{2}}{r^{3}} \times 2 \sqrt{2}$

$\Rightarrow \frac{d^{2} z}{d x^{2}}=\frac{-2 \sqrt{2}}{r}<0$

So, $z=x+y$ is maximum when $x=y=\frac{r}{\sqrt{2}}$.

Hence proved.