© Nuffield Foundation 2012

Free-Standing Mathematics Activity

Maximum and minimum

problems

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– hold as much as possible– use as little material as possible?

Manufacturers use containers of different shapes and sizes.

In this activity you will use graphs to solve such problems.

How can manufacturers design containers to:

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A drinks can must hold 330ml The manufacturer wants to find the dimensions with the minimum surface area.

Capacity330 cm3

radiusr cm

heighth cm

V = r2h

S = 2r2 + 2rh

S = 2r2 + 2πr × 2330r

S = 2r2 + r660

2330rh =

330 = r2h

Think about …Which formulae do you think will be needed to solve this problem?

Think about …How can a minimum value for S be found?

To find the minimum area, draw a graph of S against r on a spreadsheet or graphic calculator.

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0

100

200

300

400

500

600

700

0 1 2 3 4 5 6 7 8 9 10

Surf

ace

area

(cm

2 )

Radius (cm)

Surface area of a canS = 2r2 + r

660

Minimum area 260 cm2

when r = 3.7 cm

Think about…What is the minimum surface area?

Think about…How can a more accurate minimum be found?

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264.3

264.4

264.5

264.6

264.7

3.60 3.65 3.70 3.75 3.80

Surf

ace

area

(cm

2 )

Radius (cm)

Surface area of a canS = 2r2 + r

660

h =

2330r

h = 7.490 cm

Check this gives a volume of 330 cm3

Minimum S

Minimum surface area is 264.36 cm2 when r = 3.745 cm and h = 7.490 cm

= 264.36 cm2

when r = 3.745 cm

Using smaller increments of r near the minimum

= 27453π330

© Nuffield Foundation 2012

Reflect on your work • Give a brief outline of the method used to find the

minimum surface area for a can holding 330 ml of drink.

• What difference would it make to the surface area if a cuboid with square cross-section was used for holding the drink?

• Do you think a cylinder is the best shape to use? Why?

• Can you find any connections between the types of equation leading to a maximising problem, and those which lead to a minimising problem?