find the dimensions of the box
Calculus

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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.
S' = 2x  376x^2 = 0 > 2x*x^2  376x^2*x^2 = 0*x^2 > 2x^3  376 = 0
2x^3 = 376 > x^3 = 188 > x = cube root(188)
y = 94/x^2 > y = 94/(cubic root(188))^2 > y = 94/cubic root(188^2)
length = cube root(188) = 5.7286 > length = 5.73 in
width = 5.73 in
height = 94/cube root(188^2) = 2.8643 > height = 2.86 in
the height and width are wrong
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