#### Please solve rd  sharma class 12 Chapter 17  Maxima and Minima excercise 17.3  part 1 question 9  maths textbook solution

Point of local maxima value is $\frac{a}{3}$ and it’s local maximum value is $\frac{4 a^{3}}{27}.$ Also point of local minima is a and it’s local minimum value is  0.

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If $f^{\prime \prime}\left(\mathrm{c}_{1}\right)>0$ then $c_1$ is point of local minima.

If $f^{\prime \prime}\left(\mathrm{c}_{2}\right)<0$ then $c_2$ is point of local maxima .

where $c_1$ & $c_2$ are critical points.

Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.

Given:

$f(x)=x^{3}-2 a x^{2}+a^{2} x, a>0$

Explanation:

We have,

\begin{aligned} &f(x)=x^{3}-2 a x^{2}+a^{2} x \\ &f^{\prime}(x)=3 x^{2}-4 a x+a^{2} \\ &f^{\prime \prime}(x)=6 x-4 a \end{aligned}

For maxima and minima f’(x) =0

\begin{aligned} &3 x^{2}-4 a x+a^{2}=0 \\ &x=\frac{-(-4 a) \pm \sqrt{(-4 a)^{2}-4 \times 3 \times a^{2}}}{2 \times 3} \\ &\quad=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{6} \\ &=\frac{4 a \pm 2 a}{6} \\ &x=a \text { or } x=\frac{a}{3} \end{aligned}

At x= a,

\begin{aligned} f^{\prime \prime}(a) &=6(a)-4 a \\ &=2 a>0 \end{aligned}

So, x=a is point of local minima

Now at x=a,

\begin{aligned} f(a) &=a^{3}-2 a(a)^{2}+a^{2}(a) \\ &=a^{3}-2 a^{3}+a^{3}=0 \end{aligned}

Also, at $\mathrm{x}=\frac{a}{3}$

\begin{aligned} f\left(\frac{a}{3}\right) &=\left(\frac{a}{3}\right)^{2}-2 a\left(\frac{a}{3}\right)+a^{2}\left(\frac{a}{3}\right) \\ &=\frac{4 a^{3}}{27} \end{aligned}

Thus, point of local maxima is $\frac{a}{3}$& its max. value is $\frac{4 a^{3}}{27}$& point of local minima is 9 & its value is 0.