A rectangular storage container with open top is to have a volume of 50ft^3. The length of the base of the container is to be twice its width. Find dimensions that minimize cost of materials for box

Question

Material for the sides cost $5 per square foot, and materials for the base cost $8 per square foot.

Answer 1

Let the dimensions of the box be represented by LWH.

We are given L = 2W.

Since V = LWH we then have:

50 = 2W(W)H, solving for H:

H=25/W².

Now for the cost function. The area of a front is L(H) or 2W(25/W²) = 50/W. The back has the same dimensions, for for front AND back 100/W square feet. But the cost per square foot is $5, so the cost of the front and the back in terms of W is 500/W.

For the two sides.

One side has area W(H) = W(25/W²) = 25/W. Since there are two sides 50/W is there area of BOTH sides in terms of W. But each square foot costs $5, so 250/W represents the cost of the two sides.

For the base the area is 2W(W) = 2W². But each square foot costs $8. So 16W² is the cost for the base in terms of W.

The cost function looks like this:

C(W) = 500/W + 250/W + 16W² = 750/W + 16W².

When does that cost function have a minimum value? Find C'(W), set that equal to zero and solve for W to determine critical numbers (note that W must be greater than zero).

C'(W) = -750/W² + 32W

32W = 750/W²

32W³ = 750

W³ = 750/32

W= (750/32)^1/3 ≈ 2.862 feet

L = 2W ≈ 5.724

H = 25/W² ≈ 3.053

LWH = 50

Min cost: $393.11

Any other value of W greater than zero results in a greater cost.

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