The Shape of a Can

Section 1.15 The Shape of a Can

Subsection 1.15.1 Recommended Tutorials

Before starting on these exercises, you should familiarize yourself with the material covered in the following tutorials:

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Subsection 1.15.2 Introduction

Your goal in this activity is to determine the height and radius of a perfectly cylindrical can that gives a total volume of 250 mL (250 cm\(^3\)) and uses the least amount of material to create.

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Figure 1.4. A simple can in the shape of a circular cylinder.
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To construct the can, there are three surfaces: the cylindrical side and the two circles that form the top and bottom. These three surfaces will all be cut from a sheet of aluminum. The cylindrical side of the can is made from a single rectangle, so that no material is wasted from the sheet aluminum when it is cut out. However, some amount of material is always wasted from the sheet when cutting the circles for the top and bottom.

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Since it is your goal to reduce the cost of material, you will have to calculate how much of the aluminum sheet will be required to construct the can, including any material that is wasted while cutting the circles. Consider a couple possible options for cutting the circles out of a sheet of aluminum.

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Figure 1.5. Cutting circles in a square pattern.
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Figure 1.6. Cutting circles in a hexagonal pattern.
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If the circles are cut in a rectangular grid patter as in to Figure 1.5, then you can see right away just how much extra material is wasted from the sheet of aluminum. However, if you cut out circles in a hexagonal pattern as in Figure 1.6, then you have already begun to minimize the total amount of the aluminum sheet that is required for the top and bottom.

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In the exercises below, you will set up an equation for the total area of sheet metal required for the cylindrical side and the two circles, which will be cut according to this hexagonal pattern. From that equation, you will determine the height and radius of the can that minimize that area.

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Exercises 1.15.3 Exercises

1.

The side of the can can be cut from the sheet aluminum as a single rectangle, with no wasted material. One side of this rectangle has a length of \(h\) and the other side has a length equal to the circumference of the can.

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Determine an expression for the area of this rectangle in terms of its radius \(r\) and height \(h\text{,}\) and assign it to sideArea.

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Hint.

Since \(\pi\) will be involved in this expression, don’t forget to type the constant as Pi or use the palettes toolbar.

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2.

Since each of the circles will effectively use an entire hexagonal area from the sheet metal, you will need to calculate the area of a hexagon that circumscribes a circle of radius \(r\text{,}\) as shown in Figure 1.7.

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Figure 1.7. A hexagon circumscribing a circle.
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(a)

The hexagon shown in Figure 1.7 consists of six equal triangles, each with a base \(b\) and height \(r\text{.}\) Using trigonometry, determine the length of \(b\) in terms of \(r\text{.}\) Then, determine the area of one of these triangles using

\begin{equation*} A = \frac{1}{2}br\text{.} \end{equation*}

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Hint.

You may use trigonometric ratios of \(\tfrac{\pi}{3}=60^\circ\) and \(\tfrac{\pi}{6}=30^\circ\) to compute the exact length of \(b\) in terms of \(r\text{.}\)

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(b)

Determine an expression for the total area of this hexagon in terms of \(r\) only and assign it to hexagonArea.

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3.

The total amount of sheet aluminum needed for the cylindrical can is given by

\begin{equation*} totalArea = sideArea + 2 \cdot hexagonArea\text{.} \end{equation*}

Assign this expression in Maple. totalArea should be an expression that relies on both the height and radius of the can.

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4.

Since the required volume of the can is 250 cm\(^3\text{,}\) you must satisfy the constraint

\begin{equation*} 250 = \pi r^2 h\text{.} \end{equation*}

Using this constraint allows you to give totalArea as an objective function that consists only of one variable, \(r\text{.}\)

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(a)

Solve the volume constraint for \(h\) and substitute the expression into totalArea. Then, reassign this new single-variable expression to totalArea.

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(b)

Plot totalArea from part (a) over the interval r = 0 ..5, with a vertical range of A=-200..400. You should notice a minimum on the graph.

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(c)

Use the first derivative of totalArea to determine the value of \(r\) that minimizes the total area. Use the second derivative of totalArea to show that this value gives a minimum by the Second Derivative Test.

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Hint.

The minimum should occur at a critical value where the first derivative equals zero. Try using a solve() command to determine the exact value where this is true. If the second derivative at that point is negative, then the critical value must give a minimum.

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5.

Now that you know the radius \(r\) of the can, use the volume constraint to determine the height \(h\) of the can that minimizes the total amount of sheet aluminum used.

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Aside

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6.

Show that the ratio of height \(h\) to radius \(r\) is

\begin{equation*} \dfrac{h}{r} = \dfrac{4\sqrt{3}}{\pi} \approx 2.21\text{.} \end{equation*}

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