#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 10.

$\frac{25}{4}$ square units

Hint: For maxima and minima we must have $f'(x)=0$

Given: $x^2 +y^2=5^2$

Solution:

Let the base of the triangle be x, and height be y

$x^2 +y^2=5^2$

$y^2=25-x^2$

$y=\sqrt{25-x^2}$

As area of triangle A, $A=\frac{1}{2}xy$

\begin{aligned} &A=\frac{1}{2} x \times \sqrt{25-x^{2}} \\ &A(x)=\frac{x \sqrt{25-x^{2}}}{2} \\ &A^{\prime}(x)=\frac{\sqrt{25-x^{2}}}{2}+\frac{x(-2 x)}{4 \sqrt{25-x^{2}}} \end{aligned}

\begin{aligned} &=\frac{\sqrt{25-x^{2}}}{2}+\frac{x^{2}}{2 \sqrt{25-x^{2}}} \\ &=\frac{25-x^{2}-x^{2}}{2 \sqrt{25-x^{2}}} \\ &A^{\prime}(x)=\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}} \\ &A^{\prime}(x)=0 \end{aligned}

\begin{aligned} &\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}}=0 \\ &25-2 x^{2}=0 \\ &x=\frac{5}{\sqrt{2}}, y=\sqrt{25-\frac{25}{2}} \\ &x=\frac{5}{\sqrt{2}}, y=\frac{5}{\sqrt{2}} \end{aligned}

Also $A''(x)$ $=\frac{\left[-4 x \sqrt{25-x^{2}}-\frac{\left(25-2 x^{2}\right)(-2 x)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}}$

\begin{aligned} &=\frac{\left[\frac{-4 x\left(25-x^{2}\right)+\left(25-2 x^{3}\right)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}} \\ &=\frac{-100 x+4 x^{3}+25 x-2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}} \\ &=\frac{-75 x+2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}} \end{aligned}

$A^{\prime \prime}\left(\frac{5}{\sqrt{2}}\right)=\frac{-75\left(\frac{5}{\sqrt{2}}\right)+2\left(\frac{5}{\sqrt{2}}\right)^{3}}{\left(25-\left(\frac{5}{\sqrt{2}}\right)^{2}\right)^{\frac{3}{2}}}<0$

So, $x=\left (\frac{5}{\sqrt{2}} \right )$ is point of maxima

$\therefore$ Largest possible area of triangle,

$=\frac{1}{2}\left(\frac{5}{\sqrt{2}}\right)\left(\frac{5}{\sqrt{2}}\right)=\frac{25}{4}$ square units