How does this question relate to your two previous questions, let's see....
Let the base be x by x inches, and the height be y inches
V= x^2 y
27 = x^2 y
y = 27/x^2
material = 2x^2 + 4xy
= 2x^2 + 4x(27/x^2)
= 2x^2 + 108/x
d(material)/dx = 4x - 108/x^2
= 0 for a min/max of material
4x = 108/x^2
x^3 = 27
x = 3 , then y = 27/9 = 3
well, what do you know, the box must be a perfect cube of 3 by 3 by 3
A rectangular closed box with a square base is to have a capacity of 27 cubic inches determine the least amount of material required.
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