A rectangular storage container with an open top is to have a volume of 12m^3.

Question

The length of its base is twice the width. Material for the base costs $2 per square meter. Material for the sides cost $6 per square meter. Find the cost of the materials for the cheapest such container.

Answer 1

To minimize cost, you must have a cost function. Here cost (C) is a function of the surface area.

The area of the base is W•L (costs $2/m²)

The area of the two "end" sides is W•H (costs $6/m²)

The area of the two other sides is L•H (costs $6/m²)

So the cost is 2(W•L) + 6(2)(W•H + L•H)

Since "the length of its base is twice the width", we can say 2W = L so the cost equation becomes:

C = 2(W•(2W)) + 6(2)(W•H + (2W)•H)

The Volume is 12m³ and volume = L•W•H so, substituting 2W for L, we get 2W²H = 12 or H = 6/W²

So substituting "6/W2" for H in the cost equation we get cost as a function of width"

C(W) = 2(W•(2W)) + 6(2)(W•(6/W²)) + (2W)•(6/W²))

C(W) = 4W² + 12(6/W + 12/W)

C(W) = 4W² + 12(18/W)

C(W) = 4W² + 216/W

To minimize, take the derivative and set it equal to zero"

C'(W) = 8W - 216/W²

8W - 216/W² = 0

8W = 216/W²

W³ = 216/8

W³ = 27

W = 3

We are looking for the cost when W = 3 so, using our cost function:

C(W) = 4W² + 216/W

C(3) = 4(3)² + 216/(3) = 108

So the minimum cost of the material is $108