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

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

Then,

$\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=0$

$\Rightarrow 12-2 x_1=0 \Rightarrow x_1=6~~~~\left[∵ y=12 x-x^2 \Rightarrow \frac{d y}{d x}=12-2 x\right]$

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

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

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