Question 14983
Let V = volume of the smaller box.  Since {{{ V= x^3}}}, where x is the length of the side of each square side, each side of this box will be {{{x = root(3,V)}}}, and the area of each face of the box will be {{{A = x^2 = (root(3,V))^2 = V^(2/3) }}}.


Now, the larger box has twice the volume or 2V, where {{{ 2V= y^3}}}, where y is the length of the side of each square side.  Each side of this box will be {{{y = root(3,(2V))}}}, and the area of each face of the box will be {{{A = y^2 = (root(3,(2V)))^2 = (2V)^(2/3) = 2^(2/3) *V^(2/3) }}}.


Notice that the larger box has area of each face that is similar to the face of the smaller of the box, except that it is multliplied by {{{2^(2/3) }}} power.  This computes to a surface area that is about 1.587 times as much as the smaller box, so it should take about 1.587 times as long to paint the box.  Hey, it's late in Florida right now, and maybe someone will check this out and make sure I'm right!!  Any volunteers??


R^2 at SCC