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A. 4 square dm

B. 6 square dm

C. 8 square dm

D. 10 square dm

E. 12 square dm

What I've done:

\(\displaystyle x^2t=2→t=\frac{2}{x^2}\)

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:

\(\displaystyle f(x)=2x^2+4(2x)t=2x^2+8x(\frac{2}{x^2})=2x^2+\frac{16}{x}\)

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

f'(x) = 0

\(\displaystyle 4x-\frac{16}{x^2}=0\)

\(\displaystyle 4x=\frac{16}{x^2}\)

\(\displaystyle 4x^3=16\)

\(\displaystyle x^3=4\)

\(\displaystyle x=\sqrt[3]{4}\)

Minimum surface area =\(\displaystyle 2(\sqrt[3]{4})^2+\frac{16}{\sqrt[3]{4}}=2\sqrt[3]{16}+\frac{4^2}{4^{\frac13}}=2\sqrt[3]{16}+4^{\frac53}\)

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?