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Free-Standing Mathematics Activity Maximum and minimum problems

Free-Standing Mathematics Activity Maximum and minimum problems. hold as much as possible use as little material as possible?. Manufacturers use containers of different shapes and sizes. How can manufacturers design containers to:.

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Free-Standing Mathematics Activity Maximum and minimum problems

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  1. Free-Standing Mathematics ActivityMaximum and minimum problems

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

  3. radiusr cm heighth cm Capacity330 cm3 h= S= 2pr2+ A drinks can must hold 330ml The manufacturer wants to find the dimensions with the minimum surface area. 330 = pr2h V=pr2h S= 2pr2+ 2prh Think about …Which formulae do you think will be needed to solve this problem? S= 2pr2+ 2πr × Think about … How can a minimum value for S be found? To find the minimum area, draw a graph of S against ron a spreadsheet or graphic calculator.

  4. 260 cm2 Minimum area when r=3.7 cm S = 2pr2+ Think about… How can a more accurate minimum be found? Think about… What is the minimum surface area?

  5. = 264.36 cm2 when r= 3.745 cm h = = S = 2pr2+ Using smaller increments of r near the minimum MinimumS h = 7.490 cm Check this gives a volume of 330 cm3 Minimum surface area is 264.36 cm2when r = 3.745 cm andh= 7.490 cm

  6. 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?

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