OPEN AND CLOSED PROBLEM SOLVING

A mathematical investigation can be very focussed or quite broad depending on whether the problem set is ‘open’ or ‘closed’. Consider a closed problem with a fixed answer – few assumptions need to be drawn in building the model. A typical closed problem might be to determine the water capacity held in a tank, which has a circular cross-section of three metres and a depth of 4 metres. We can immediately use our knowledge of cylindrical volumes to determine the capacity of the tank. (The author was actually set this problem by a farmer once!)

Other times detailed results are called for and much data is needed as an input for the model. In fact we often build models from data where mathematical relationships can be found without a clear knowledge of the underlying cause. This is especially true with probabilistic models and those which draw on statistics.

For an open problem we may need to list a number of conditions which apply so as to create a suitable model with real application. When testing the model these assumptions may need to be refined.

A typical example of an open problem related to the same mathematics as our tank problem is the following.

We can see that the volume of a can will depend on the height and the radius.

We also observe that the amount of metal used to make the can will depend on the surface area of the can.

We can make a first approximation to this problem by making some assumptions.

1. The cost of making a steel can is the cost of the material used in making it.

Why are there are no real giants?

E

^XAMPLE3.15

S

o l u t i o n

Many things we buy are stored and sold in steel cans – what is the ‘best’

height and radius of such cans?

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^XAMPLE3.16

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o l u t i o n

h cm r cm

91

2. The can is the shape of a cylinder and metal used is only that given by that geometry.

These two assumptions can be questioned but they give us a starting point. We begin with the basic geometry implied by assumption 2.

The volume, V cubic centimetres of the can is given by where the radius is r and the height is h in centimetres.

If we unfold a can we see that it is made up of two circles (the top and bottom) and a rectangle making the curved surface. The surface area, A of the can is seen to be

Generally we decide in advance how much is to be sold – that is – the volume is fixed. We can now see that if we do fix the volume of the can then the radius and height of the can are directly related and that as the radius increases the height decreases and vice versa. We also know that V varies for different things. A soft drink can holds typically between 325 and 375 ml (each millilitre is a cubic centimetre) whereas paint may be sold in containers of between 1 litre and 4 litres. But when we know the value of V we can determine h if we know r for example.

Since h and r are both measured in the same units it will be useful to look at the ratio h/r, which will be dimensionless.

Let us simplify the problem by looking at a 1 litre container and set V = 1. (This actually has the effect of measuring h and r in decimetres but we will not use actual lengths, just their ratio.)

Now, from we have

and so, after substituting into the equation for the surface area , we have

Giving us a simple equation, showing how the area varies with the can radius for a fixed volume.

We can easily graph this using our calculators and show that it has a unique minimum. The calculator allows us to do this numerically and then we can find the value of h and the ratio.

In fact it is possible (using advanced mathematics called calculus) to show that this occurs when V = πr²h

h cm 2πr cm

cm² A = 2πr²+2πrh

V = πr²h 1 πr²h h 1 πr²

---⇒ =

=

h 1

πr²

---= A = 2πr²+2πrh

A 2πr² 2πr 1 πr²

---× +

=

2πr² 2 πr ---+

=

A

r

92

and that this gives . We can actually show also that this ratio holds regardless of the volume of the steel can. If we use a general V instead of V = 1 we get the same ratio.

So far we have converted our open problem into a very closed problem using our assumptions.

We could immediately look at some real steel cans and see what ratios we actually find. In practice we find is far more common. In fact, it may often exceed 3! Does this mean the manufacturers like to waste money?

We now open up the problem by looking at the assumptions. We assume that the metal used is all in the can geometry. But the circular ends at least are stamped from sheets, which leave a lot of waste.

Each circle may be stamped from a square of side and leaves waste . Then this means that the actual metal used really has an area of

From this we can review our analysis and show that (we can check this by using our graphing calculators) and so we find that (approximately). In fact, theory shows that this gives . This is larger than our previous value and shows that the waste metal generated is an issue.

It may be inflating the ratio too much however, despite the fact that real observed ratios are generally larger. We can look at this stamping process more closely. We imagined that the circles were stamped out of metal sheets in squares as shown but they are more efficiently stamped if the process is offset to stamp them out of hexagons which can also cover the metal sheet (with some edge effects we can neglect overall).

r 1

2π

---= 3 h

r--- = 2

h ---r>2

waste

2r 4r²–πr²

A 8r²+2πrh 8r² 2 ---r +

= =

r 1

2

---= h

---r = 2.55 h

r--- 8 π

---=

This can reduce the ratio again by cutting waste. We look at the simple geometry involved. Each regular hexagon can be seen to comprise 6 equilateral triangles as shown in the diagram. These triangles have side length d where, by Pythagoras’s theorem, d² r² 1

2---d

  ²∴d

+ 2r

3

---= =

r d

93

Each triangle has area and there are 6 triangles used to cover the hexagon used to stamp out the circle. The area stamped is then . We use two of these in each can and so the rule for the steel can surface area is now

We again graph this using our calculator and estimate the values of r and hence h. We see this ratio is approximately 2.21. It is possible to show that this is exactly but we need not prove this here.

This is less than before so we have saved on waste metal. What else have we neglected?

We could look at the problem in more detail. In practice the steel can must have circles stamped of larger radius than the can radius to enable a ‘lip’ to be produced to fold the ends over and connect them to the can side. This increases the metal usage and so the cost. The actual

production process will also include this sealing up of the can lips and there are also likely to be fixed costs associated with steel can production runs anyway. It is possible to see that these require a modification of our assumptions and that they suffice to increase the h/r ratio.

So far we note that this problem has only used simple geometry and graphing to get numerical estimates of the ratios. Many real problems will call on higher mathematics like calculus to get results as we are commonly dealing with rates of change or averages.

If we look closely at the real situation for the steel can dimensions problem we would find that in practice the optimal ratio will depend on volume V - for larger cans we find that the ratio is generally lower and nearer our present theory. The reason for this depends on sealing costs, but it is not important to see why this is so at present … only to note that we can make increasingly more complex models by changing assumptions in an open problem in order to get more realistic results. It is clear that even our simplest model gave some idea of the real situation however.

We see from this example that a mathematical investigation may lead to open problem solving where answers depend on assumptions and we often need to compare theory with observation to improve model representation. From this example we can revisit our general rules about

modelling and mathematical investigation.

1. Formulate the problem.

We first need to decide what we want to know.

2. Outline a model.

Identify input and output variables and their relationships. Set up equations.

3. Examine model for usefulness.

Is the model as built able to be used. This means can we get the input variables and other data needed and can we solve the equations at all or possibly within a fixed time frame which we may have (some models need to be run on tight timelines – a typical example may be a scheduling or rostering system)

4. Test the Model.

Does the output predicted by the model match the results observed.

1 2---dr

2 3r²

A 4 3r²+2πrh 4 3r² 2 ---r +

= =

h

---r 4 3

---π

94 5. Refine the Model.

If the answer to step 4 is NO we may need to refine the model and move back to steps 3 and 4.

There are many ‘closed’ problems in Chapter 23 that, like the example we have just completed, can be transformed into an open ended problem. Many of which are ideal for your portfolio work.

The Binomial Theorem – CHAPTER

4

95

In document IB Maths SL Textbook (Page 106-111)