Answer
$4\times4\times2\ ft$
Work Step by Step
Step 1. Assume the edge length of the square base is $x\ ft$ and the height is $y\ ft$; the volume is given as $x^2y=32\ ft^3$ which leads to $y=\frac{32}{x^2}$
Step 2. The goal is to find the minimum surface area to achieve the least weight with a fixed thickness. The total surface area is given by $A=x^2+4xy=x^2+\frac{128}{x}$
Step 3. Take the derivative to get $A'=2x-\frac{128}{x^2}$; we see that $A'=0$ leads to $x^3=64$ which gives $x=4\ ft$. Thus the dimensions should be $4\times4\times2\ ft$
Step 4. Check $A''=2+\frac{256}{x^3}\gt0$; we confirm the above dimensions give the minimum surface area.
