Optimize a soda can. Imagine you are an engineer for a soda company.

Question

Imagine that you are an engineer for a soda company, and you are tasked with finding the most economical shape for its aluminum cans. You are given a set of constraints: the can ust hold a volume V of liquid and be a cylindrical shape of height h and radius r, and you need to minimize the cost of the metal required to make the can.

(a) First, ignore any waste material that is discarded during the manufacturing process and just minimuze the total surface area for a given volume V. Using this constraint, show that the optimal dimensions are achieved when h = 2t.

(b) Next, take the manufacturing process into account. Materials for the cans are fut from flat sheets of metal. The cylindrical sides ar made form curved rectangles, and rectangles can be cut from sheets of metal with virtually no waste. However, when the disks of the top and bottom of the can are cut from flat sheets of metal, there is significant waste material. Assume that the disks are cut from squares with side lengths of 2r, so that one disk is cut out of each square in a grid. Show that in this case the amount of material is minimized when: h÷r = 8÷π ≈2.55 .

(c) It is far more efficient to cut the disks from a tiling of hexagons than from a tiling of squares, as the former leaves far less waste material. Show that if the disks for the lids and bases of the cans are cut from a tiling of hexagons, the optimal ratio is h ÷ r = 4√3 ÷ π ≈ 2.21 . Hint: The formula for the area of a hexagon circumscribing a circle of radius r is A = 6r ^2 ÷ √3.

(d) Look at a variety of aluminum cans of different sizes from the supermarket. Which models from problems a-c best approximate the shapes of the cans? Are the cans actually perfect cylinders? Are there other assumptions about the maufacture of the cans that we should take into account? Do a little bit of research, and write a response to answer some of these questions by comparing our models to the actual dimensions used.

Answer 1

We will need the volume formula for a cylinder to answer the question:

1. V = πr²h

We are told to fix V and vary only h and r. However, in doing so we can express h as a function of r by isolating h in the volume formula. This will be helpful for the rest of the problem. The resulting function is:

1. h = V/(πr²)

(a) To solve this part, we need only minimize the surface area A of the cylinder. The surface area is given by

1. A = 2πr² + 2πrh

However, we have expressed h as a function of r, so we may substitute this expression into the surface area formula:

1. A = 2πr² + 2πr*V/(πr²) = 2πr² + 2V/r

If we are to minimize this, we should take the derivative with respect to r and set the obtained expression equal to 0. Our goal is to find where the slope is 0, which could indicate that the function has "bottomed out."

1. dA/dr = 4πr - 2V/(r²) = 0
2. ⇒ 4πr³ - 2V = 0 (multiplying by r² is allowed since we know that r ≠ 0)
3. ⇒ r³ = V/(2π)
4. ⇒ r = (V/(2π))^1/3

The second derivative of A with respect to r, that is, d²A/dr², is positive, indicating that the curve A is concave up. As a result, the obtained value for r is, indeed, a local minimum!

We can now pass this value of r into our function h, giving us

1. h = V/(π*(V/(2π))^2/3)
2. = 2^2/3 * (V/π)^1/3
3. = 2 * (V/(2π))^1/3
4. = 2r,

as desired.

(b) Now we are constrained by a different area formula, since the total area of material used is no longer just the surface area of the cylinder. The curved rectangular side still has area 2πrh, but the top and bottom of the can now each require an area of material given by (2r)² = 4r². This gives us the following area formula:

1. A = 8r² + 2πrh

Substituting the expression we obtained for h in terms of r:

1. A = 8r² + 2V/r

We now differentiate with respect to r and find the roots.

1. dA/dr = 16r - 2V/r² = 0
2. ⇒ 16r³ - 2V = 0
3. ⇒ r³ = V/8
4. ⇒ r = (V^1/3)/2

Again, A is concave up here, so this is a local minimum.

To finish, we make one modification to the steps we used in part (a). Instead of directly passing the obtained value for r into our function h, we will manipulate the equation slightly so that we can immediately get the required ratio h/r

1. h = V/(πr²)
2. ⇒ h/r = V/(πr³) (accomplished by dividing both sides by r)

Now we substitute our expression for r on the right side only:

1. h/r = V/(π*((V^1/3)/2)³)
2. = V/(π*V/8)
3. = 8/π,

as desired.

Note that the last step would have been easier if we had stopped at r³ when solving for the root of dA/dr. That way we could have immediately plugged that into the denominator instead of going through the extra step of cubing r. We will do this in part (c), which again asks us to solve for h/r.

(c) We can follow the same steps as in (b). We are told the relevant hexagonal area is 6r²/√3, so our new area formula is:

1. A = 12r²/√3 + 2πrh

Substituting for h yields:

1. A = 12r²/√3 + 2V/r

Differentiating with respect to r and setting the expression equal to 0 brings us to:

1. dA/dr = 24r/√3 - 2V/r² = 0
2. ⇒ 24r³/√3 - 2V = 0
3. ⇒ 24r³/√3 = 2V
4. ⇒ r³ = V*(√3)/12

We stop here because, as explained previously, it makes our lives easier.

Substituting this into the previously obtained expression for h/r, we get:

1. h/r = V/(π*V*(√3)/12)
2. = 12/(π√3)
3. = 12(√3)/(3π) (obtained by multiplying numerator and denominator by √3)
4. = 4(√3)/π

as desired.

(d) Can Manufacturers Institute (http://www.cancentral.com/beverage-cans/standards) lists the standard beverage can dimensions as 2.602 × 4.812 (diameter × height), in inches. To get the h/r value, we have to divide the diameter in half, then take height/radius. We get a value of h/r ≈ 3.699, which is fairly far from all of the computed values. The closest is the can from part (b), but this is still far off.

To account for the discrepancy, we note that there are a number of other considerations that come into designing a can. First, soda cans are not perfect cylinders. They have to be designed in a way that actually permits drinking the beverage without spilling it or cutting the consumer. Second, soda cans have to be ergonomic. That is, the average consumer should be able to hold the can in one hand without the can slipping. Third, soda cans also have to be designed to allow for easy storage and shipping. Also, soda cans must be structurally sound. Here's a fantastic video which concerns the design of soda cans: https://www.youtube.com/watch?v=hUhisi2FBuw