Hello acacia,
I would orient the box such that the width is $x$, the height is $y$ and the length is $z$. Hence, the bottom panel has area $xz$, the side panels have a total area of $2yz$ and the front and back panels have a combined area of $2xy$.
Let all linear measures be given in feet.
Thus, our objective function, the function we wish to minimize is the cost function in dollars, which is given by:
$$C(x,y,z)=3xz+yz+2xy$$
Subject to the constraint on the volume:
$$g(x,y,z)=xyz-6=0$$
Using Lagrange multipliers, we obtain:
$$3z+2y=\lambda(yz)$$
$$z+2x=\lambda(xz)$$
$$3x+y=\lambda(xy)$$
Solving for $\lambda$, the first two equations imply:
$$\frac{2y+3z}{yz}=\frac{2x+z}{xz}$$
Cross-multiplying, we obtain:
$$2xyz+3xz^2=2xyz+yz^2$$
$$3xz^2=yz^2$$
Since the constraint requires $$0<z$$, we may write:
$$3x=y$$
In like manner the first and third equations above imply:
$$\frac{2y+3z}{yz}=\frac{3x+y}{xy}$$
Cross-multiplying, we obtain:
$$2xy^2+3xyz=3xyz+y^2z$$
$$2x=z$$
Substituting for $y$ and $z$ into the constraint, we obtain:
$$x(3x)(2x)=6$$
$$x^3=1$$
$$x=1\implies y=3,\,z=2$$
Observing that:
$$C(1,3,2)=3(1)(2)+(3)(2)+2(1)(3)=18$$
and another constraint value such as $(x,y,z)=(1,2,3)$ yields:
$$C(1,2,3)=3(1)(3)+(2)(3)+2(1)(2)=19$$
We may then conclude:
$$C_{\min}=C(1,3,2)=18$$