# Calc 1 (Optomization)

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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.

• Calc 1 (Optomization) -

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

• Calc 1 - correction -

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)

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