Question

A box manufacturer is to make a closed box of volume 288in3, where the base is a rectangle whose length is three times its width. What is the length of the box that uses the least amount of material

Answer 1

V = Length x Width x Height = 288
If we call the width w
then length = 3w
And height = 288/(3w x w)

Assuming none of the material used overlaps the box will need 6 faces ( top, bottom, 2 ends and 2 sides)
Area of material = 2 x LW + 2 x HW + 2 x HL
= 2(3w x w + 288/(3w x w) x w + 288/(3w x w) x 3w)
= 2(3w^2 + 288/3w + 288/w)
= 2(3w^2 + 96/w + 288/w)
= 2(3w^2 + 384/w)
= 6w^2 + 768/w
Taking the derivative in terms of w
A’ = 12w - 768/w^2
A’ is zero when 12w = 768/w^2
w^3 = 768/12 = 64
w = cube root of 64 = 4
This makes the box with the minimum materials 12 x 4 x 6
So the length that uses the minimum material is 12 inches