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

2. © Nuffield Foundation 2012 – 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:

3. © Nuffield Foundation 2012 A drinks can must hold 330ml The manufacturer wants to find the dimensions with the minimum surface area. Capacity 330 cm 3 radius r cm height h cm V = r2hV = r2h S = 2  r 2 + 2  rh S = 2  r 2 + 2πr × S = 2  r 2 + h =h = 330 =  r 2 h 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.

4. © Nuffield Foundation 2012 S = 2  r 2 + Minimum area 260 cm 2 when r = 3.7 cm Think about… What is the minimum surface area? Think about… How can a more accurate minimum be found?

5. © Nuffield Foundation 2012 S = 2  r 2 + h = h = 7.490 cm Check this gives a volume of 330 cm 3 Minimum S Minimum surface area is 264.36 cm 2 when r = 3.745 cm and h = 7.490 cm = 264.36 cm 2 when r = 3.745 cm Using smaller increments of r near the minimum =

6. © 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?