I have to find $k$ to maximize the volume of the parallelepiped spanned by the vectors $A = (6,k,0), B = (3,4,-k)$ and $C = (1,2,2)$.
I know the volume would just be $f(x)= ABC$ and that I take the triple scalar product to get the parallelepiped spanned by those vectors, but would I use LaGrange to maximize?
HINT
Compute volume as the triple scalar product, as a function of $k$, to get $V(k)$, and then since $V: \mathbb{R} \to \mathbb{R}$ you can use ordinary calculus.
Moreover, since $V$ is a low-degree polynomial in $k$, basic algebra could be enough :)
UPDATE
In view of the comment, you get $$ \begin{split} V(k) &= (6,k,0) \cdot (3,4,-k) \times (1,2,2) \\ &= \det \begin{bmatrix} 6 & k & 0 \\ 3 & 4 & -k \\ 1 & 2 & 2 \end{bmatrix} \\ &= 6 \det \begin{bmatrix} 4 & -k \\ 2 & 2 \end{bmatrix} - k \det \begin{bmatrix} 3 & -k \\ 1 & 2 \end{bmatrix}\\ &= 6(8+2k)-k(6+k)\\ &= 48 + 12k - 6k - k^2 \\ &= 48 + 6k -k^2 \end{split} $$ which is a basic parabola facing down, so the maximum occurs at $$k = \frac{-b}{2a} = \frac{-6}{2 \cdot (-1)} = 3$$ with maximum volume of $$V(3) = 48 + 18-9=57.$$