Question

Applications of Derivative.

An open box with a capacity of 36000 cubic inches is needed. IF the box must be twice as long as it is wide, what dimensions would require the least amount of material?

Answer 1

Volume= 36,000
Let x be width, y is height, 2x is length
Goal is to minimize surface area
SA = (x*2x) + 2(x*y) +2(2x*y)
SA = 2x^2 + 6xy
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Volume = x*2x*y = 2x^2y
2x^2y = 36,000
y = 18,000/x^2
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Substitute into SA equation
SA = 2x^2 + 6x(18,000/x^2)
SA = 2x^2 + 108,000/x
Differentiate and set equal to 0
dA/dx = 4x -108,000/x^2 = 0
4x = 108,000/x^2
4x^3 = 108,000
x^3 = 27,000
x = 30
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Solve for y
y = 18,000/30^2
y = 20
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Dimensions which minimize surface area are
width = 30
length = 60
height = 20