Answer
$ y= \frac{1}{3}$ is the minimum, when $ x=\frac{1}{6}$
Work Step by Step
Given $$ y=4x-12x^{2} $$
So, we have
\begin{aligned} y&=4x-12x^{2} \\
&=-12\left(x^{2}-\frac{1}{3} x\right)\\
&=-12\left(x^{2}-\frac{1}{3} x+\frac{1}{36}-\frac{1}{36}\right)\\
&=-12\left((x-\frac{1}{6} )-\frac{1}{36}\right)\\
&=-12\left((x-\frac{1}{6} )^2-\frac{1}{36}\right)\\
\end{aligned}
The lowest value for the squared term is $0$. So, we see that $ y=\frac{12}{36}=\frac{1}{3}$ is the minimum, when $ x=\frac{1}{6}$.
