Optimization problems as far as I can remember are almost always solved by the fact that the maximum or minimum of a function is at a point where the derivative equals zero, i.e. where the slope is equal to zero. Therefore the strategy becomes.. what is the thing (e.g. volume, line distance, etc) we want to maximize/minimize, simplify it down to one variable and then we take the derivative with respect to that variable and set it equal to zero. Then we solve for that variable.

1) I'm guessing this is a square piece of material from which we are to cut a plus-shaped looking shape that can be folded up into an open top box. If not, then there is a different solution.

Solution #1: I like to keep everything written as a variable because it makes it much easier to check your solution after you are done. So, I define:

l = length of one side of what I assume is a square piece of material
w = width of the box's base
d = depth of box's base
h = height of the box

The thing we want to maximize is the volume. Volume in the variables we defined is w*d*h... the problem states that it has a square base, so w = d.. and we can write it as V = w^2*h. If we cut the plus pattern out of a square piece of material then l = 2h + w. Since l is a constant, we want an expression for volume that has only one variable... so we solve for w and plug that in to our volume expression.

w = l - 2h =>
V = (l - 2h)^2*h
V = (l^2 - 4*l*h - 4*h^2)*h
V = l^2*h - 4*l*h^2 - 4*h^3

So this is the expression for which we want to know at what value of h is it a maximum... so we take the derivative and set it equal to zero. (if you are uncomfortable taking a derivative with respect to h you could insert x for h.. but you should get comfy with taking the derivative with respect to any variable...)

dV/dh = l^2 - 8*l*h - 12*h^2

This equation now defines the slope of the function V as function h... we want to know when that slope equals zero, and then we solve for h... at this point I'll plug in the value for l, which I assume is the square root of 1200...

1200 - 8*sqrt(1200)*h - 12*h^2 = 0

Which when I solve gives me roots of 3.7 and -26.8... which has got to be an incorrect answer i think... hmm... shitty... Anybody see my mistake?... i'm fairly confident that my solution is appropriate.

The 2nd way to do it would be to figure out how one can make a box of maximum value with a surface area of exactly 1200 cm^2... something like 4*w*h+w^2 = 1200 for the area, were w^2*h is volume... solve the top equation:
(1200 - w^2)/4w = h;
plug into volume equation:
V = w^2*(1200 - w^2)/4w
V = 300w - (w^3)/4
and do the same thing as above....

dV/dw = 300 - 3/4*w^2 = 0
w = sqrt(-300*(-4/3))
w = 20 cm

Solve for h in the surface area equation:

h = (1200 - w^2)/4w
h = (1200 - (20^2)/(4*20)
h = 10 cm

check answer... 10cm * 20 cm = 200 cm^2*4 = 800 cm^2, and 20 cm*20cm = 400 cm^2... for a total surface area of 1200 cm^2.

The volume associated with that box is 400cm^2*10cm = 4000 cm^3.

Still bothered by my first solution... they should give you different answers.. but I still doubt that it is correct. Thoughts?