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:

[tex]x^2t=2→t=\frac{2}{x^2}[/tex]

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:

[tex]f(x)=2x^2+4(2x)t=2x^2+8x(\frac{2}{x^2})=2x^2+\frac{16}{x}[/tex]

To determine the value of x so that the surface area will be minimum:

f'(x) = 0

[tex]4x-\frac{16}{x^2}=0[/tex]

[tex]4x=\frac{16}{x^2}[/tex]

[tex]4x^3=16[/tex]

[tex]x^3=4[/tex]

[tex]x=\sqrt[3]{4}[/tex]

Minimum surface area =[tex]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}[/tex]

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?