#### Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 18.

$A=r^2$

Hint: For maxima or minima f'(x) must be zero f'(x) = 0

Given:  $\frac{x^2}{4}+y^2=r^2$

Solution: $\frac{x^2}{4}+y^2=r^2$

$x^2 +4y^2=4r^2$

$x^2 =4(r^2 -y^2)$                ......(1)

Area = xy

Square on both sides

$A^2=x^2 y^2$

$Z= 4y^2(r^2-y^2)$                ......from (1)

Now,

\begin{aligned} &\frac{d Z}{d y}=8 y r^{2}-16 y^{3} \\ &\frac{d Z}{d y}=0 \\ &8 y r^{2}-16 y^{3}=0 \\ &8 r^{2}=16 y^{2} \\ &y^{2}=\frac{r^{2}}{2}, y=\frac{r}{\sqrt{2}} \end{aligned}

Substitute y value in eqn (1)

\begin{aligned} &x^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right) \\ &=4\left(r^{2}-\frac{r^{2}}{2}\right) \\ &=4\left(\frac{r^{2}}{2}\right) \end{aligned}

$x^2 = 2r^2$

$x= r\sqrt{2}$

\begin{aligned} &\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48 y^{2} \\ &\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48\left(\frac{r^{2}}{2}\right) \\ &=-16 r^{2}<0 \end{aligned}

Thus the area is maximum when $x=r\sqrt{2}$ and $y= \frac{r}{\sqrt{2}}$

Area = xy

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

$A=r ^2$