Question 769328
As written, the problem is unclear. Perimeter is applied to 2-dimensional figures. Perimeter is the distance all around a 2-dimensional figure.  The word perimeter can not be applied to the 3-dimansional box. It could be applied to the square base of the box or to the rectangular side of the box.
 
{{{x}}}= length of a side of the square base of the box
{{{h}}}= height of the box
{{{4x}}}= perimeter of the base of the box
{{{2x+2h}}}= perimeter of the side of the box
{{{x^2}}}= area of the base of the box
{{{(x^2)h}}}= volume of the box
 
The height of the box being the same as the perimeter of the side of the box,
{{{h=2x+2h}}},
does not make sense, because {{{x}}} and {{{h}}} must be both positive inch measurements.
 
It must be that the height of the box is the same as the perimeter of the base of the box,
{{{h=4x}}}
Then the volume of the box is
{{{(x^2)(4x)=108in^2}}}
Solving:
{{{(x^2)(4x)=108in^2}}}-->{{{4x^3=108in^2}}}-->{{{x^3=108in^2/4}}}-->{{{x^3=27in^2}}}-->{{{highlight(x=3in)}}}
Then, the height of the box, {{{h=4x}}}, is
{{{h=4(3in)}}}-->{{{highlight(h=12in)}}}