# Mathematics Project on Efficiency in Packaging

## Efficiency in Packaging

Hexagonal Packing

Efficiency in packin

Objective

To investigate the efficiency of packing of objects of different shapes in a cuboid box.

Efficiency is the percentage of box space occupied by the objects.

Description

1. Took a certain number of cylindrical tins and packed them in a cuboid container.

**(a) **For illustration I took 81 tins.

**(b) **Second time I took 64 tins.

**(c) **Third time I took 49 tins.

**2. **The cylindrical tins can be placed in two different ways These are

**(a) **Square packing

**(b) **Hexagonal packing

**3. **I wished to study which packing out of two is more efficient.

4. To understand the difference between the two packing I have drawn figures on the left side pages.

Example 1 (81 tins) Calculation

Case 1

Square packing

Each base circle is circumscribed by a square. Area of one circle = n R^{2}

Area of square = 4 R^{2}

Area of circle / area of square = nR^{2} / 4 R^{2}

This ratio will be evidently the same as the cross section of all the tins to the total base area.

Percentage efficiency = n / 4 x 100

= 78.5 %

Therefore the efficiency in case of square packing is 78.5%

Case 2

Hexagonal packing

Here we determine the sides of the base of the container in terms of the radius of the cylindrical tin.

One side of the rectangular base i.e. BC = 18 x R.

To determine the other side, AB = 2 x R + 9 x h, where h is the altitude of the equilateral triangle formed by joining the centres of three adjacent circles.

h = 2R sin 60°

AB = 2R + 18 R sin 60°

sin 60° = V3 / 2

Now AB = 2R + 18R x V3 / 2 = (2 + 9V3) R

Area of ABCD = 18R x (2 + 9V3) R = 18R^{2} (2 + 9V3)

Percentage efficiency = 81nR^{2} x100 / 18R^{2} (2 + 9V3)

= 80.3 %

Example 2 (64 tins) Calculation

Case 1

Square packing

Each base circle is circumscribed by a square. Area of one circle = n R^{2}

Area of square = 4 R^{2}

Area of circle / area of square = nR^{2} / 4 R^{2}

This ratio will be evidently the same as the cross section of all the tins to the total base area.

Percentage efficiency = n / 4 x 100

= 78.5 %

Therefore the efficiency in case of square packing is 78.5%

Case 2

Hexagonal packing

Here we determine the sides of the base of the container in terms of the radius of the cylindrical tin.

One side of the rectangular base i.e. BC = 16 x R.

To determine the other side, AB = 2 x R + 8 x h, where h is the altitude of the equilateral triangle formed by joining the centres of three adjacent circles.

h = 2R sin 60°

AB = 2R + 16 R sin 60°

but sin 60° = V3 / 2

Now AB = 2R + 16R x V3 / 2 = 2R + 8RV3 = (1 + 4V3) 2R

Area of ABCD = 16R x (1 + 4V3) 2R = 32R^{2} (1 + 4V3)

Percentage efficiency = 64nR^{2} x100 / 32R^{2} (1 + 4V3)

= 79.3 %

Example 3 (49 tins) Calculation

Case 1

Square packing

Each base circle is circumscribed by a square. Area of one circle = n R^{2}

Area of square = 4 R^{2}

Area of circle / area of square = nR^{2} / 4 R^{2}

This ratio will be evidently the same as the cross section of all the tins to the total base area.

Percentage efficiency = n / 4 x 100

= 78.5 %

Therefore the efficiency in case of square packing is 78.5%

Case 2

Hexagonal packing

Here we determine the sides of the base of the container in terms of the radius of the cylindrical tin.

One side of the rectangular base i.e. BC = 14 x R.

To determine the other side, AB = 2 x R + 7 x h, where h is the altitude of the equilateral triangle formed by joining the centres of three adjacent circles.

h = 2R sin 60°

AB = 2R + 14 R sin 60°

but sin 60° = V3 / 2

Now AB = 2R + 14R x V3 / 2 = 2R + 7RV3 = (2 + 7V3) R

Area of ABCD = 14R x (2 + 7V3) R = 14R^{2} (2 + 7V3)

Percentage efficiency = 49nR^{2} x100 / 14R^{2} (2 + 7V3)

= 77.96 %

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Remarks

**1. **In the calculations here the number of tins was fixed and the cuboid dimensions are variable.

A similar exercise may be done with fixed cuboid dimensions and variable number of tins.

**2. **We can also determine the efficiency for packing of spheres in a cuboid.

Volume of sphere = 4/3 nR^{3}

Volume of cube = 8R^{3}

Percentage efficiency = 4 n R^{3} / 3 x 8 R^{3}

= n / 6 = 52 %

Note

When 81 and 64 tins were taken Hexagonal packing was more efficient but in case of 49 tins Square packing was more efficient.