#### Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 28 maths.

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

Hint: For maximum or minimum values of z, we must have $\frac{dz}{dx}=0$

Given:

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

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

Solution:

Now,

$z=x+y$

$z=x+\sqrt{r^2-x^2}$                            ......from (1)

\begin{aligned} &\frac{d z}{d x}=1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}} \\ &\frac{d z}{d x}=0 \\ &1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}=0 \\ &2 x=2 \sqrt{r^{2}-x^{2}} \\ &x=\sqrt{r^{2}-x^{2}} \end{aligned}

Squaring on the both sides

$x^2=r^2-x^2$

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

Substituting the value of x in equation (1)

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

\begin{aligned} &y=\sqrt{r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}} \\ &\frac{d^{2} z}{d x^{2}}=\frac{-\sqrt{r^{2}-x^{2}}+x(-x)}{\sqrt{r^{2}-x^{2}}}=\frac{-r^{2}+x^{2}-x^{2}}{r^{2}-x^{2}} \end{aligned}

$=\frac{-r^2}{r^3}\times 2 \sqrt{2}$

$=-\frac{ 2 \sqrt{2}}{r}<0$

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