A rectangular storage container with an open top is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of the materials for the cheapest such container.

Answer: $81.77

Step-by-step explanation:

The area of the base is W * 2W = 2W^2

The volume of the container = Base area * height ... which implies that

10 = 2W^2 * H ⇒ H = 10 / [ 2W^2 ] = 5 / W^2

So ... the total surface area is given by ... area of the base + side area =

2W^2 + 2 (5/W^2) [ W + 2W] =

2W^2 + (10/W^2}[ 3W] =

2^W^2 + 30/W

So ... the cost, C, to be minimized is this:

C = (base cost of materials) (base area) + (side cost of materials) (side area)

C = 5 (2W^2) + 3 (30/W)

C = 10W^2 + 90/W

Note: Calculus ... take the derivative of the cost ... set to 0 and solve

So we have

C' = 20W - 90/W^2 = 0

20W = 90/W^2

W^3 = 90/20

W^3 = 9/2

W^3 = 4.5

W = (4.5) ^ (1/3) ≈ 1.651

Subbing this back into the cost function, the minimized cost is

C = 10[ (4.5) ^ (1/3) ] ^2 + 90/[4.5]^ (1/3) ≈ $81.77