Calculus (3rd Edition)

$y= \frac{1}{3}$ is the minimum, when $x=\frac{1}{6}$
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 $x=\frac{1}{6}$.