The open box has a square base. Let the dimensions of the base be X. Let the height of the box be H.

As the surface area of the open box is C, we get:

X^2 + 4*X*H = C

=> H = (C - X^2)/4X

The volume of the box is X^2*(C - X^2)/4X

=> V = `(X*(C - X^2))/4`

To determine the dimensions that maximize C, differentiate V with respect to X

`(dV)/(dX) =(X*(-2X) + (C - X^2))/4`

=> `(-2*X^2 + C - X^2)/4`

=> `(-3*X^2 + C)/4`

Now solve dV/dX = 0 for X

`(-3*X^2 + C)/4 = 0`

=> `X^2 = C/3`

=> `X = sqrt(C/3)`

H = (C - X^2)/4X = `(C - C/3)/(4*sqrt(C/3)) = sqrt(C/3)/2`

**To maximize volume, the base should have dimensions `sqrt(C/3)` and the height should be `sqrt(C/3)/2`**