Suppose the box is to be of volume $V_0$ cubic cm; and the cost of material for the front and back sides is $b$ dollars per square cm, $c$ dollars per square cm for the left and right two sides, and $d$ dollars per square cm for the top and bottom sides. Note that $V_0$, $b$, $c$ and $d$ are just arbitrary constants.
This is the task and so far all I've been able to get together is the following equation:
$$f(x,y,z)=2bxy+2cyz+2dxz$$
$$g(x,y,z)=xyz=V_0$$
$$\overrightarrow {\nabla}f = f_x + f_y + f_z$$
$$f_x = 2by+2dz$$
$$f_y = 2bx + 2cz$$
$$f_z = 2cy + 2dx$$
$$\overrightarrow {\nabla}g = g_x+g_y+g_z$$
$$g_x=yz = \lambda yz$$
$$g_y=xz = \lambda xz$$
$$g_z=xy = \lambda xy$$
So then I set the corresponding functions equal to each other and get:
$$2by+2dz=\lambda yz \;;\frac{2by+2dz}{\lambda} = yz$$
$$2bx+2cz=\lambda xz \;;\frac{2bx+2cz}{\lambda} = xz$$
$$2cy+2dx=\lambda xy \;;\frac{2cy+2dx}{\lambda} = xy$$
But I'm lost with where to go from here. Any hints or tips?
when constrained to the surface $xyz=V_0$
$f(x,y,z)= 2V_0(\frac{b}{z}+\frac{c}{x}+\frac{d}{y})$ so $\vec \nabla f(x, y, z) =-2V_0(\frac{c}{x^2}, \frac{d}{y^2},\frac{b}{z^2} ) $
$g(x, y, z) = xzy \\ \implies \vec \nabla g(x, y, z) = (yx, xz, xy)$
the gradient constrained to the surface is $\vec \nabla g(x, y, z) = V_0(\frac 1x, \frac 1y, \frac 1z) $
So $$ \frac{-2V_0c }{x^2}=\frac{\lambda V_0}x \implies x= \frac{-2c}{\lambda} \\y= \frac{-2d}{\lambda} \\z= \frac{-2b}{\lambda} \\ \text{ and } \frac{-8bcd}{\lambda^3}=V_0 \implies \lambda = \left( \frac{-8bcd}{V_0} \right)^\frac 13$$