# If the solve the problem

Question:

$\mathrm{f}(\mathrm{x})=x \sqrt{1-x}, x>0$

Solution:

Given : $f(x)=x \sqrt{1-x}$

$\Rightarrow f^{\prime}(x)=\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}}=\frac{2-3 x}{2 \sqrt{1-x}}$

For the local maxima or minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0$

$\Rightarrow x=\frac{2}{3}$

Since, $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{2}{3}, x=\frac{2}{3}$ is a point of maxima.

The local maximum value of $f(x)$ at $x=\frac{2}{3}$ is given by

$\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9}$