Find the maximum value of box?

What's to be maximised? The volume of a cuboid with a square base and a height.
the base material is given with a square sheet of metal that has sides of length 8cm.

The relation between the side length of the base b and the height h is obviously:
b = 8 - 2h
So, the cuboid will have the volume
v = h * (8 - 2h)^2.

Our target function is then:
f (x) = x (8 - 2x)^2

The problem-related domain of the function is determined by the set of sensible values for x which we feed into it: A height of h <= 0 would make as little sense as a value of
h >= 4. Thus we are only interested in the values in the closed interval [0, 4].
We can write this:

D = { x | 0 < x < 4 }.

We want to know, where this function has its maximum value in the problem-related domain. We, therefore have to find the roots of the 1st derivative:

f '(x) = 2x (8 - 2x) * -2 + (8 - 2x)^2
f '(x) = -4x (8 - 2x) + (8 - 2x)^2
f '(x) = (8 - 2x) ( -4x + 8 - 2x)
f '(x) = (8 - 2x) ( -6x + 8)

The 1st derivative is zero, where f (x) has its extrema. So,

(8 - 2x) ( -6x + 8) = 0 ;; divide by 2
(4 - x) ( -3x + 4) = 0

gives us at a glance:
x1 = 4, x2 = 4/3.

x1 goes away (as this would leave us with 4 little square pieces of 4cm side length each, but no base for the box, which is why we excluded it from the domain in the first place).

Is there at f (x2) a maximum or a minimum? The second derivative will tell us:

f ''(x) = -3 (4 - x) + (-1)( -3x + 4)
f ''(x) = 6x - 16

and

f ''(4/3) = -8 < 0 --> Maximum!

Thus the side length of the square pieces that have to be removed from each corner is

h = 4/3 cm = 1.33 cm, the side length of the base is
b = (8 - 2.66) cm = 5.34 cm, and, finally, the volume
v = 1.33 * 5.34^2 = 37.93 cm^3.