Approach to Optimisation Problems

1. Draw a picture to represent the situation2. Annotate with information you know & variables3. Write equations with given info4. 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 value7. Answer in context (units)

OPTIMISATIONExample 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 24 000 mm3, 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=60o 4x+3y=60o 3y=60-4xo Y=20-4x/3

• Area=2xy • Area=2x(20-4x/3)• Area=40x-8x2/3

• 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=150m2

x

yy

x

y

xx

Example 2: A block of ice is shaped like a cuboid. The volume is 24 000 mm3, 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=24000mm3

• 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’=0+36-48000x-2=0• 48000/x2=36• X2=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=5295.73mm2=5296mm2

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.

20

0)2(400

0400800

0800400

0800400'

400100400

400100

400Pr

400

1$100$

2

32

32

32

21

22

2

orx

xx

xx

xx

xxP

xxP

xx

xofit

xopleNumberofpey

yCost