Imagine you are an engineer for a soda company, and you must find the most economical shape for its aluminum cans. You are given this set of constraints. The can must 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 discarded during the manufacturing process and just minimize the total surface area for a given volume, V. Using this constraint, show that the optimal dimensions are achieved when h = 2r. The formula for the volume of a cylinder is V = πr 2h. The formula for the lateral area of a cylinder is L = 2πrh. b) Next, consider the manufacturing process. Materials for the cans are cut from flat sheets of metal. The cylindrical sides are made from curved rectangles, and rectangles can be cut from sheets of metal leaving virtually no waste material. However, the process of cutting disks for the tops and bottoms of the cans from flat sheets of metal leaves 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 needed 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 for different-sized aluminum cans 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 manufacture of the cans that we should consider? Do a little bit of research, and write a one-page response to answer some of these questions by comparing our models to the actual dimensions used.

Final answer:

The optimal dimensions of a soda can to minimize surface area given a specified volume are achieved with the ratio h = 2r. This changes to h/r = 8/π or 2.55 when considering rectangles, and h/r = 4√3/π or 2.21 when considering hexagons. Real-world soda cans typically use dimensions somewhat between these models.

Explanation:

Given the volume V of a cylindrical can with height h and radius r, we can find the dimensions that minimize the surface area. The volume of a cylinder is given by the formula V = πr2h. To minimize the surface area, which is given by 2πrh + 2πr2, we should find the derivative of this with respect to r and set it equal to 0, which gives us h = 2r.

When considering waste materials from cylinders cut from a sheet of metal, this modifies the equation for the lateral surface area. If the disks for the ends of the cans leave waste, and we consider that each disk is cut from a square of side length 2r, the optimal dimensions change to h/r = 8/π ≈ 2.55.

If the ends are cut from a tiling of hexagons, the area of a hexagon can be written as A = 3√3r2/2. This minimizes the waste material and results in the optimal ratio h/r = 4√3/π ≈ 2.21.

Comparing these models to actual can dimensions, we find that real-world can dimensions usually fall somewhere between the cylindrical and the hexagonal model, leaning slightly more toward the cylindrical.

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Answer:

a) To minimize the total surface area for a given volume, V, we need to consider the formula for the lateral surface area of a cylinder:

Now, set this derivative equal to zero:

2πr = 0

This implies that r can be any positive value, and it doesn't affect the minimization of TSA. Therefore, we can choose any value for r. However, it's more convenient to work with h = 2r because it simplifies the calculations and is often used in the design of cans.

b) When considering the manufacturing process with minimal waste, the total material needed is minimized when the ratio h/r = 8/π ≈ 2.55. This is because when h/r is equal to 8/π, you can fit the largest number of disks with a diameter of 2r onto a square sheet of metal with side lengths of 2r, minimizing waste.

c) 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. Hexagons are more efficient than squares for packing circles, which represents the lids and bases of the cans. This results in less waste material.

d) To compare these theoretical models to actual aluminum cans from the supermarket, you would need to measure the dimensions of real cans and calculate their ratios of h/r. While many aluminum cans are close to the idealized cylinder shape, real-world manufacturing considerations can lead to variations. For instance, cans might have slightly different ratios of h/r based on manufacturing efficiencies, branding, and design choices. Additionally, the exact shapes and dimensions of cans may vary among different brands and beverage types.

It's also important to note that real cans might have additional features such as ridges, embossing, or variations in the shape of the top and bottom, which can affect their overall dimensions and shape. Therefore, it's essential to consider these factors when comparing theoretical models to actual cans.