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Let's set that the sides of the square base of the box is x (length and width is x) and the height of the box is y. So then the volume of the box would be the volume of a cube which is:

V = (length)(width)(height) -------> V = (x)(x)(y) ----------> V = (x^2)(y)

On the other hand, in order to use a minimum amount of material then we just need to minimize its surface area which is:

S = Area of the base + 4*(Area of each wall)

The area of the base is a square and the area of each wall would be rectangles with length x and width y. So we would have:

S = x^2 + 4*(x)(y) ------------> S = x^2 + 4xy

We solve for y from the equation of the volume and by the way let's remember v = 94 in^3. So we would have:

V = (x^2)(y) ------> 94 = (x^2)(y) ----------> 94/x^2 = (x^2)(y)/x^2 ----------> y = 94/x^2

Then we enter it into the surface area equation:

S = x^2 + 4xy ----------> S = x^2 + 4x(94/x^2) --------------> S = x^2 + 376/x -----------> S = x^2 + 376x^-1

Then we find its derivative, equal it to zero, and solve for x.