FreeStanding Mathematics Activity Maximum and minimum problems Nuffield

Free-Standing Mathematics Activity Maximum and minimum problems © Nuffield Foundation 2012

Manufacturers use containers of different shapes and sizes. How can manufacturers design containers to: – hold as much as possible – use as little material as possible? In this activity you will use graphs to solve such problems. © Nuffield Foundation 2012

A drinks can must hold 330 ml radius r cm The manufacturer wants to find the dimensions with the minimum surface area. Capacity height 330 cm 3 h cm 330 = pr 2 h V = pr 2 h h= S = 2 pr 2 + 2 prh S = 2 pr 2 + 2πr × S = 2 pr 2 + 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. © Nuffield Foundation 2012

S = 2 p r 2 + Minimum area 260 cm 2 when r = 3. 7 cm Think about… What is the minimum surface area? © Nuffield Foundation 2012 Think about… How can a more accurate minimum be found?

Using smaller increments of r near the minimum S = 2 pr 2 + Minimum S = 264. 36 cm 2 when r = 3. 745 cm h= = h = 7. 490 cm Minimum surface area is 264. 36 cm 2 when r = 3. 745 cm and h = 7. 490 cm © Nuffield Foundation 2012 Check this gives a volume of 330 cm 3

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? © Nuffield Foundation 2012