Approach to Optimisation Problems 1.Draw a picture to represent the situation 2.Annotate with information you know & variables 3.Write equations with given.

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Presentation transcript:

Approach to Optimisation Problems 1.Draw a picture to represent the situation 2.Annotate with information you know & variables 3.Write equations with given info 4.Write expression to be min/maximised in terms of one variable only (usually requires use of another equation). 5.Differentiate & make f’(x)=0 6.Solve for the unknown value 7.Answer in context (units)

OPTIMISATION Example 1: Two adjoining rectangular yards share a boundary. There is 60 m of fencing available for the boundaries. Calculate the maximum total area for the two yards. Example 2: A block of ice is shaped like a cuboid. The volume is mm 3, and the depth is 18 mm. The ice takes longer to melt if the surface area is as small as possible. Calculate the minimum surface area for this block. Example 3: The cost of running a swimming pool is $100 per day plus $1 per swimmer who uses the pool. The number of people prepared to pay $x to use the pool can be approximated by the formula. Calculate the price of entry which maximises the profit for the pool operators.

Example 1: Two adjoining rectangular yards share a boundary. There is 60 m of fencing available for the boundaries. Calculate the maximum total area for the two yards. Perimeter=60 o 4x+3y=60 o 3y=60-4x o Y=20-4x/3 Area=2xy Area=2x(20-4x/3) Area=40x-8x 2 /3 Area’=40-16x/3=0 16x/3=40 16x=120 X=7.5m Y=20-(4*7.5)/3 Y=10 The maximum total area is 2*7.5*10=150m 2 x y y x y x x

Example 2: A block of ice is shaped like a cuboid. The volume is mm 3, and the depth is 18 mm. The ice takes longer to melt if the surface area is as small as possible. Calculate the minimum surface area for this block. V=24000mm 3 V=18xy=24000 y=(24000/18x) SA=2xy+36x+36y SA=2x(24000/18x)+36x+36(24000/18x) SA=(48000/18)+36x+(48000x -1 ) SA’= x -2 = /x 2 =36 X 2 =48000/36 x=√(48000/36) X=36.51mm (2dp) Y=36.51mm(2dp) Min SA= SA=(48000/18)+36*36.51+(48000/36.51) min SA= mm 2 =5296mm 2 18mm y x

Example 3: The cost of running a swimming pool is $100 per day plus $1 per swimmer who uses the pool. The number of people prepared to pay $x to use the pool can be approximated by the formula. Calculate the price of entry which maximises the profit for the pool operators.