##### find the dimensions of the box

label Calculus
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Jul 16th, 2015

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)

Jul 16th, 2015

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

Jul 16th, 2015

the height and width are wrong

Jul 16th, 2015

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Jul 16th, 2015
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Jul 16th, 2015
Oct 24th, 2017
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