The dimensions of the box are (S-2x)by(S-2x)byS = 2S/3 by 2S/3 S by S/6 to obtain a box with largest possible volume
In physics and mathematics, the dimension of a mathematical space (or object) is informally defined as the minimum number of coordinates needed to specify any point within it. Thus, a line has a dimension of one (1D) because only one coordinate is needed to specify a point on it – for example, the point at 5 on a number line. A surface, such as the boundary of a cylinder or sphere, has a dimension of two (2D) because two coordinates are needed to specify a point on it – for example, both a latitude and longitude are required to locate a point on the surface of a sphere. A two-dimensional Euclidean space is a two-dimensional space on the plane. The inside of a cube, a cylinder or a sphere is three-dimensional (3D) because three coordinates are needed to locate a point within these spaces.
Let S = length of the sides of the square cardboard sheet. Let x = the side of each square cut from the corners. The volume, V, of the box when the sheet is folded up is:
V = (S-2x)(S-2x)x = 4x3 - 4Sx2 + S2x
To find the maximum value, take the derivative of V wrt x, set it to zero, and solve for x:
V ' = dV/dx = 12x2 - 8Sx + S2
0 = 12x2 - 8Sx + S2
0 = (6x-S)(2x-S)
x = S/6 and S/2
To determine which is the max and which is the min, take the second derivative of V wrt x.
V'' = d2V/dx2 = 24x - 8S
At x = S/2, V'' = 24(S/2) - 8s = 4S. Since S>0, the point x=S/2 is a minimum. This makes sense because you have cut away the entire cardboard square.
At x = S/6, V'' = 24(S/6) - 8S = -4S < 0, so x = S/6 is the maximum
The dimensions of the box are (S-2x)by(S-2x)byS = 2S/3 by 2S/3 S by S/6
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