Question

If an open box has a square base and a volume of 111 in.3 and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.)

Answer 1

6.06 inches square by 3.03 inches high

Explanation:

An open-top box will have a minimum surface area when its lateral area is double its base area. That is, left-area + right-area = front-area + back area = bottom area.

With that in mind, the simplest way to compute the dimensions is to find the base edge length as the cube root of double the volume.

b = ∛(2×111 in³) ≈ 6.06 in

h = b/2 = 3.03 in . . . . . there are 4 faces; each is half the area of the base

The box is 6.06 inches square and 3.03 inches high.

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Solution using derivatives

Let b represent the edge length of the base. The volume is ...

V = b^2h

Then the height is ...

h = V/b^2

h = 111/b^2

and the total surface area is ...

A = b^2 + 4bh = b^2 +4·111/b

The area is maximized when the derivative is zero:

dA/db = 0 = 2b -444/b^2

Solving for b, we get ...

2b^3 = 444

b^3 = 222

b = ∛222 ≈ 6.06 . . . . . . as above