Asked by gillian
A rectangular closed box with a square base is to have a capacity of 27 cubic inches determine the least amount of material required.
Answers
Answered by
Reiny
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
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
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