What Is the Minimum Surface Area for a Cuboid Box with a Square Base?

[Monoxdifly]

The volume of a cuboid box with a square base is 2 litres. The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides. Suppose the side length of its base is x and the height of the cuboid is h. The minimum production cost is reached when the surface area is ...
A. 4 square dm
B. 6 square dm
C. 8 square dm
D. 10 square dm
E. 12 square dm

What I've done:
$$x^2t=2→t=\frac{2}{x^2}$$
The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides. I tried to put that into account regarding the surface area f(x), and thus:
$$f(x)=2x^2+4(2x)t=2x^2+8x(\frac{2}{x^2})=2x^2+\frac{16}{x}$$
To determine the value of x so that the surface area will be minimum:
f'(x) = 0
$$4x-\frac{16}{x^2}=0$$
$$4x=\frac{16}{x^2}$$
$$4x^3=16$$
$$x^3=4$$
$$x=\sqrt[3]{4}$$
Minimum surface area =$$2(\sqrt[3]{4})^2+\frac{16}{\sqrt[3]{4}}=2\sqrt[3]{16}+\frac{4^2}{4^{\frac13}}=2\sqrt[3]{16}+4^{\frac53}$$

I'm stuck. Pretty sure I misinterpreted the whole "The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides." thing. What was I supposed to do?

 

[skeeter]

surface area, $A = 2x^2 + 4xh$

$h = \dfrac{2}{x^2} \implies A = 2x^2 + \dfrac{8}{x}$

cost of the top and bottom is twice that of the lateral sides ...

$C = k \left(4x^2 + \dfrac{8}{x} \right)$, where $k$ is a production cost constant

$C’ = k \left(8x - \dfrac{8}{x^2} \right) = 0 \implies x = 1$

 

[Monoxdifly]

Why is the 2 only multiplied by $$2x^2$$ and not with $$\frac8x$$?

 

[skeeter]

Why is the 2 only multiplied by $$2x^2$$ and not with $$\frac8x$$?

top & bottom, surface area = $2x^2$, costs twice as much to produce than the lateral sides, surface area = $4xh = 4x \cdot \dfrac{2}{x^2} = \dfrac{8}{x}$

 

[Monoxdifly]

Ah. Right. That makes sense. Thank you.