The box has a square base and has a open top. Let the side of the square base be S and the height of the box be H.
The surface area of the box is S^2 + 4*S*H.
S^2 + 4*S*H = 1200
=> H = 300/S - S/4
The volume of the box is V = S^2*H = S^2*(300/S - S/4)
=> V = 300*S - S^3/4
To maximize V, solve `(dV)/(dS) = 0`
=> 300 - (3*S^2)/4 = 0
=> (3*S^2)/4 = 300
=> S^2 = 400
=> S = 20
`(d^2V)/(dV^2) = -3/4*(2*S)` which is negative for S = 20. This shows that the maximum volume of the box is when S = 20.
H = 300/S - S/4 = 15 - 5 = 10
The maximum volume is 4000 cm^3.
The dimensions for the maximum volume are 20x20x10 which gives a volume of 4000 cm^3.