Maximizing volume in Calculus

Question

An open box is to be created from a flat piece of cardboard $36$ inches square by cutting a square from each corner and then folding up the edges. How long should the side of the square being cut out in order to maximize the volume of the box created, and what is the volume of the box created?
So far I have $f(x)=x(6-2x)^2$. Would $f'(x)=2x(6-3x)$ and the critical points being $x=0$ and $x=3$?

Answer 1

If the side of the square you're cutting out is $x$ inches, then the box created will have dimensions, in inches, $(6-2x) \times (6-2x) \times x$. Hence, you need to maximize $f(x) = x(6-2x)^2$, where $0 \leq x \leq 3$.