Person who understands math I choose you!!!!!

You have a picture.....use that as your sketch

2.volume = l x w x w = lw^2     (w= side of the square)

3. 54000 cu in   is max

4. l + 4w =180   or  l= 180-4w   substitute this in to #2

    (180-4w)(w^2) = volume   

       (180w^2 - 4w^3)   a 'local' max will be where slope = 0

      Derivative  360 w - 12w^2  = 0

                          w = 0 or 30    (throw out '0' obviously)

         w=30  then l = 180-4w=60 

volume max = l x w x w = 60 x 30 x 30 = 54000 cu in

1. It looks like the picture pretty much shows the cross-section as being a square

2.  We have that  G + L   =  108  ⇒  G  =  108 - L

Assuming  that the  cross-section is a square....its side is 1/4 of the girth and can be expressed as:

G/4  =  (108 - L) / 4

So....the volume  can be expressed as   L [ (108 - L) / 4] ^2

So we have that

V  = L [ (108 - L) / 4]^2   =  L [ 27  - L/4]^2 = L [ L^2/16 - 27L/2  + 729 ]  =

L^3/16  - 27L^2//2  +  729L

To find the max volume, we can use some Calculus  or a graph

The graph seems easiest....here it is :

https://www.desmos.com/calculator/ybfbga97jl

3.  Looking at the graph, the maximum volume is [ again, assuming a square base ]  = 11664  in^3

4.  The Length, L  =  36 in

The Girth, G   =  108  - 36   =   72 in

So....the side of the square  is  1/4 of this =  18 in

So...the dimensions are   18 in x 18 in x 36 in   =  11664 in^3

Note that the restrictions have been met   Girth + Length  =  (4 * 18)  +  36   =   72 + 36  =  108 in

Thanks to you all so much!!!