Minimizing Packaging Costs A rectangular box is to have a square base and a volume of $20f{t}^3.$ If the material for the base costs $$\frac{0\mathrm{.}20}{f}{t}^2,$ the material for the sides costs $$\frac{0\mathrm{.}06}{f}{t}^2,$ and the material for the top costs $$\frac{0\mathrm{.}10}{f}{t}^2,$ determine the dimensions $($in ft $)$ of the box that can be constructed at minimum cost. $($Refer to the figure below.$)$

$($i$)$

$x=$

$$

$y=$ ft