What do the glas sizes have to be on a aquuarim that is 160 liters so you use the least amount of glas and there isn't a glas on top. So no glas on top.

160L  = 160,000cm^3

Not too sure about this one....but I believe that an edge of ∛160,000  ≈ 54.288 cm  in each dimension will minimize the cost......total area of the glass  =   5 * (∛160,000)^2  ≈  14,736.126 cm^2

[ You forgot to convert liters to cm^3......your configuraion uses  4000 + 3200 + 8000  = 15.200cm^2  of glass ]

This lessens  the required glass in your configuration by about 464cm^2

However....aquariums are not usually designed as cubes...  [ maybe because of too much water pressure on the sides??  ]

Maybe two of our other members  - Alan or Heureka - know of the most efficient [workable] design......???

I would say that a cube would minimize your glass surface area.....if you use tempered glass you can make a cube no problem...

dimensions would be   x * x * x = 160000    x = 54.28 cm  and you would need 5 of them (no top)

5 pieces of  54.28 x 54.28  glass   (probably 3/16 (maybe 1/8th)temepered glass ...or acrylic))

I believe most aquariums are not cubes because they typically want to maximize the viewing from the FRONT, so the sides are usually smaller        I once had a 40 gallon one that was octagon shaped.

Thanks, EP......!!!

I suspect that your reasoning for aquarium design is correct.....!!!

Hmm!

Let W = width, L = length, H = height

V = WLH                 A = WL + 2HL + 2WH

(i) W = L = H

V = W³            A = 5W² 

(ii)  L = 2^1/2W     H = 2^-1/2W

V = W* 2^1/2W*2^-1/2W    so   V = W³ still.

A = W*2^1/2W + 2* 2^-1/2W*2^1/2W + 2*W*2^-1/2W    so  A = (2^1/2 + 2 + 2^1/2)W²

A = 2(1 + 2^1/2)W²     or  A ≈ 4.828W²  i.e. a slightly smaller area for the same volume!

Thanks, Alan......!!!!