Question

A rectangular storage container with an open top is to have a volume of k cubic meters. The length of its base is twice its width. The material for the base costs $6 per square meters and the material for the sides costs $10 per square meter.

Find the dimensions of the container having the least cost.

Answer 1

area of base = 2w^2
height = k/2w^2
area of sides = 4*k/2w^2 = 2k/w^2

cost = 6(2w^2) + 10(2k/w^2)
= 12w^2 + 20k/w^2

dc/dw = 24w - 40k/w
dc/dw = 0 when

24w = 40k/w
24w^2 = 40k
w^2 = 5/3 k
w = √(5k/3)

least cost box is thus √(5k/3) x √(5k/3) x 3/10

Answer 2

error starts here:

dc/dw = 24w - 40k/w

It should be:

dc/dw = 24w - 40k/w^3
24w = 40k/w^3
24w^4 = 40k
w = ∜(5k/3)

least cost box is thus ∜(5k/3) x ∜(5k/3) x √(3k)/(2√5)

Answer 3

Umm, I thought we were computing a "Rectangular" box? Shouldn't that 2k/w^2 be a 3k/w?

Area of sides =2 (2wh + wh)

Answer 4

= 2 (3wh)
= 6wh
= 6w (k/2w^2)
= 6wk/2w^2
= 3k/2w

Answer 5

= 3k/w