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let length and width of bas be x (its square so both are equal)
and height be y.
so volume $\displaystyle x^2y = 2$
so$\displaystyle y = 10/x^2$
area of sides and bottom = $\displaystyle x^2 + 4xy$
area of top =$\displaystyle x^2$
so cost
$\displaystyle C = 10(x^2 + 4xy) + 20x^2$
$\displaystyle = 30x^2 + 40xy$
$\displaystyle = 30x^2 + 400/x$ (since $\displaystyle y = 10/x^2)$
for minimum cost
dC/dt = 0
solve for x now.
then solve for y ($\displaystyle y= 10/x^2$)