Hint: So we want to minimize $4x+2y+z$ subject to the condition that $x$, $y$. and $z$ are positive and $xyz=27$.
If you have already seen Lagrange Multipliers, this is a good place to use the method. If you have not, see the alternate approach below. Let $F(x,y,z,\lambda)=4x+2y+z+\lambda(xyz-27),$ and set the partial derivatives of $F$ with respect to $x$, $y$, $z$, and $\lambda$ equal to $0$. Solve and see what this tells you. You will need to check that there really is a minimum at the point you find.
Another way: We have $z=\frac{27}{xy}$. So we want to minimize $G(x,y)$, where $G(x,y)=4x+2y+\frac{27}{xy}.$ Calculate the partial derivatives of $x$ and $y$ with respect to $x$ and $y$, and set them equal to $0$. Solve. Then you have to show that we really do get a minimum.
Remark: You are doing this for a calculus course, so I have suggested calculus ways. But there is (in my opinion) a nicer way. Let $u=4x$, $v=2y$, and $w=z$. We want to minimize $u+v+w$ subject to the condition that the variables are positive, and $uvw=(27)(8)$.
By the Arithmetic Mean Geometric Mean Inequality (known affectionately as AM-GM to generations of Math Olympiad contestants) we have $\frac{u+v+w}{3}\ge (uvw)^{1/3},$ with equality only when $u=v=w$.
So $\frac{u+v+w}{3}\ge 6$, with equality when $u=v=w$. That gives minimum value of $u+v+w$ equal to $18$. Putting $u=v=w$ we find that each is equal to $6$, so our minimum is reached at $x=\frac{3}{2}$, $y=3$, and $z=6$.
The above remark is intended to show that calculus is not the only tool for such problems. It will also serve as a check for your Lagrange Multiplier (or other) calculation.