The point on the curve $y=12x-x^2$ where the tangent is parallel to x-axis, is

Solution:

Let $\left(x_1, y_1\right)$ be the required point on $y=12x-x^2$

Then,

$\begin{array}{l}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=0\\ \Rightarrow 12-2x_1=0\Rightarrow x_1=6\left[\because y=12x-x^2\Rightarrow \frac{dy}{dx}=12-2x\right]\end{array}$

Since $\left(x_1, y_1\right)$ lies on $y=12x-x^2$

$\therefore y_1=12x_1-{x_1}^2\Rightarrow y_1=72-36=36$

Hence,$( 6, 36)$ is the required point