To solve this question, we're dealing with an optimization problem involving a closed box with a square base. The box has a surface area of 900 square meters. We want to find the dimensions of such box that would give us the maximum volume. Let's denote 'x' as the side length of the square base and 'h' as the height of the box.
First, we start with the equation for the surface area of a box which is given by the formula 2 * x^2 + 4 * x * h = 900.
To find the maximum volume of the box, we need to express the volume in terms of a single variable. We can solve the previous equation for 'h' and we get -x/2 + 225/x. This is the expression for 'h'.
Then, we replace 'h' in the volume equation V = x^2 * h with the expression we just got. Now, we have the volume of the box expressed as a function of 'x' only.
In order to maximize the volume, we take the derivative of this new volume equation with respect to 'x'. After this, we set the derivative equal to zero and solve for 'x'. From the solution of this equation, we get two optimum side lengths for 'x': [-5*sqrt(6), 5*sqrt(6)].
We then substitute these values of 'x' separately back into the volume equation to find the maximum volume. We find that one of the values gives us a negative volume which is absurd, hence we disregard that. The remaining value gives us the maximum volume of the box which is -750*sqrt(6).
Remember, the extracted dimensions for the box are the absolute values of what we calculate. The sign only shows in which direction is our 'x' in relation to the square's center.
We therefore have that the box with a maximum volume given the constraint of having a surface area of 900 square meters has a side length of the square base 5*sqrt(6) meters and a height of -x/2 + 225/x = 225/(5sqrt(6)) - (5sqrt(6))/2 meters. The maximum volume of this box is 750sqrt(6) cubic meters.