MODELLING – PHASE 2

There are no real changes in moving from Phase 1 to Phase 2. The changes, where they exist, are rather subtle. They might involve the use of diagrams, an aspect that did not feature in Phase 1.

They might involve the use of trial and error or the construction of a function as opposed to an equation.

For example, if the problem is referring to the area of a square, then after defining the side length of the square as having a length x cm, you might decide that it is appropriate to intoduce the function to define the area of this square.

One difference, however, can occur in the way information is provided and therefore, how that information and/or data is used. Here is an example:

A common task that is peformed by farmers is setting up enclosures on their farm. A particular farmer has 30 metres of fencing with which to enclose an area of 100 . The farmer uses a wall to act as one side of the area to be enclosed. What are the possible dimensions of this enclosure?

We begin by using a diagram that includes the information provided:

Next we introduce variables to define the length and width, so,

let the length be y m and the width x m.

Then, as we are given 30 metres of fencing, we must have that

i.e., – (1) Also, we are told what the area is to be, 100 .

As the area is a feature of this problem we also define it by introducing a function to represent the area:

Let the area enclosed be defined by , so that .

Now, at this point in time we observe that there are three variables, two independent ones, x and y and the dependent variable A.

Somehow we need to reduce this to one independent variable and one dependent variable. We keep A and now look to keep one of x or y. The choice as to which one is kept is not important (in this example), so we choose to keep x, meaning that we need an expression for y in terms of x.

From (1) we have that .

This means that we can now express A in terms of one variable, namely

However, we know that the area must be 100 , therefore,

That is, we now have a quadratic equation to solve.

A x( ) cm²

m²

x metres y metres

wall

x+y+x = 30 2x+y = 30 m²

A m² A = x×y

2x+y = 30⇔y = 30–2x

A = x×y = x×(30–2x) m²

100 = x 30( –2x) 100 = 30x–2x²

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When x = 10, the length is 30 – 2(10) = 10, giving an area of . √ When x = 5, the length is 30 – 2(5) = 20, giving an area of . √ It is always a good idea to check your answer.

So that the possible dimension satisfying the restictions are:10 m by 10 m and 5 m by 20 m Comparing this solution with those in the previous section, we see that there is a build-up to the solution, in the sense that we had three variables to start with and then somehow had to express one in terms of the other. Although we did finish by solving an equation, we first created a function. Now, these are subtle differences, but differences nonetheless. This does not mean that all problems in this section will have three variables to start with. Each question will need to be dealt with on its own merit.

Before moving onto some examples, lets consider a slight alteration to the question above.

This time, we want:

Of all possible rectangular enclosures, which one will have the maximum area?

Of course we could use the method of trial and error. Some possible scenarioes are show below.

As we can see, different dimensions produce different areas (in some cases, the same area!).

To determine the dimensions that will provide the largest area we make use of the area function, . In fact, because A is a function of x only, we can write it as

All that remains is to sketch the graph and use it to determine its maximum value.

Notice that if x = 0, A = 0 and if x = 15, A = 0.

So, we need to restrict the values of x to 0 < x < 15.

Our definition of A now becomes, , 0 < x < 15.

Using the CALC menu and option 4:maximum we were able to obtain the maximum area as 112.5 .

2x²–30x+100 = 0

⇔2(x–5)(x–10) = 0

∴x = 5 or x = 10

10×10 = 100 m² 5×20 = 100 m²

8 m

11m 11m

4m 4m

22 m 8m 8m 14 m

A = 88 m² A = 88 m² A = 112 m²

A = x 30( –2x)

A x( ) = x 30( –2x)

A x( ) = x 30( –2x)

0 < x < 15 m²

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Unlike previous problems, we are supplied with all the variables and the equation!

In fact, there is no need to define or introduce any new variables. However, here is a problems where our task is to interpret the information that has been given.

We start by visualising and interpreting the situation:

As t represents time, then we know that t ≥ 0.

Also, when t = 0, we have that

This tells us that initially, the ball is 1m above ground level (the point of projection).

Having obtained some ‘feel’ for what is happening, we now start answering the questions.

(a) We wish to determine t when h = 16, so, we solve the equation

As we are asked for the first time, then t = 1. That is, it took 1 sec to reach 16 m.

But what of t = 3, what does it represent?

As the ball must come down, t = 1 represents the time when the ball is 16 m above ground level on its way up. While t = 3 represents the time taken for the ball to be 16 m from the ground, but this time it is on its way back down. This also means that it took 2 seconds for the ball come back to a height of 16 m from the time it first reached that height on its way up.

(b) We recognise the equation as a quadratic, so we can sketch its graph and use it to determine the maximum height by locating its turning point.

Using the CALC menu and option 4:maximum we have that the maximum height reached is 21 metres, 2 seconds after it was projected.

(c) When h = 0, the ball will have reached ground level and so, we need to find that value of t for which h = 0.

i.e.,

There are a number of options that can be used.

e.g., use CALC menu and select option 2:zero or the TRACE option or as we have done, use solve from the catalogue menu.

Giving t = 4.05 (to 2 d.pl.).

The height, h m, above ground level, reached by a ball, t seconds from when it is thrown, is given by the equation .

(a) How long did it take for the ball to first reach a height of 16 metres?

(b) What is the maximum height reached by the ball?

(c) For how long is the ball in flight?

h = 20t–5t²+1

E

^XAMPLE3.6

S

o l u t i o n

ground level point ofprojection h ball will reach a maximum

height at some time.

ball on the way up

ball on the way down, eventually reaching the ground

h = 20×0–5×0²+1 = 1

16 = 20t–5t²+1⇔5t²–20t+15 = 0

⇔5(t²–4t+3) = 0 5 t( –3)(t–1)

⇔ = 0

∴t = 3 or t = 1

0 = 20t–5t²+1

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Had we used the quadratic formula in part (c), we would have obtained the following solutions,

However, as t ≥ 0 we would have rejected t = –0.0494.

We find that linear models are used in many facets of life as was seen in Chapter 6 and will be seen in Chapter 24. One area where this is very common is that of economics and business.

Examples can be found in equations that model company profits and revenue as well as the costs inccured during the running of a business. There are a number of terms that are often referred to when dealing with economic and business situations. Some of these terms are; profit, revenue, cost, depreciation, market demand, market supply, market equilibrium and break-even point.

Analysing economic behaviour through mathematical interpretation is a skill which is sought by many of the leading firms around the world. Modelling such economic behaviour displays the power inherent in mathematics.

(a) When n = 1000, C = 12(1000) + 40000 = 52000

That is, it will cost the manufacturer $52,000.

(b) Let R, dollars, denote the revenue. If the manufacturer has sold 1000 calculators, the revenue is given by R = 32 × 1000 = 32000.

Therefore the manufacturer makes a profit of $32,000 – $52,000 = – $20,000.

A negative profit means that the manufacturer has made a loss.

Therefore the manufacturer has lost $20,000!

(c) The break-even point occurs when no gain or loss is made, that is, when R = C.

Therefore, if n calculators are produced, we have

That is, to break-even, the manufacturer needs to produce (and sell) 2000 calculators.

Although we have answered the questions, it will be of great use if we can visualise this situation 0 20t–5t²+1∴t –20± 20²–4×–5×1

2×–5

---= =

4.0494,–0.0494

=

The total cost, $C, to a manufacturer of calculators is given by the equation where n is the total number of calculators produced. The manufacturer sells these calculators to suppliers at a fixed price of $32 each.

(a) Find the manufacturer’s cost when 1000 calculators are produced.

(b) Will the manufacturer make a profit when 1000 calculators are produced?

(c) How many calculators must this manufacturer produce in order to break-even?

C = 12n+40000

E

^XAMPLE3.7

S

o l u t i o n

Because the profit is defined as the revenue – cost, i.e., P = R – C, the break-even point occurs when P = 0.

32n = 12n+40000

⇔20n = 40000

⇔n = 2000

69 by sketching the relationships involved:

From the graphs, we see that the break-even point corresponds to the point of intersection of the two straight lines representing

A. the cost function, and

B. the revenue function,

Notice how we have only sketched the graphs for values of n that are greater than (or equal to) zero.

The above example illustrates that the amount of items a manufacturer sells and the price of the item determine the total amount of money earned by selling the items. If the item is more expensive the number sold commonly decreases. The cost of producing items often depends on a fixed (set-up) cost and a price per item (resource cost). Three other terms that are of importance in economic modelling are:

Market demand The quantity of a product (goods or services) that consumers are willing to buy at various prices.

Market supply The quantity of a product (goods or services) that producers are prepared to offer for sale at various prices.

Market equilibrium This occurs when the quantity demanded equasl the quantity supplied.

(a) The revenue is .

Suppose that we have found that x units can be sold daily for a price of p dollars per unit where and the cost of making these x units is given by the function

Find (a) the daily revenue function, . (b) the daily profit function .

Assuming the production capacity is at most 400 units per day, find (c) how many units should be made and sold to maximize profit.

(d) the maximum profit per day.

(e) the unit price to be charged to maximize the profit.

x = 800–p

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(c) The profit function is given by a quadratic function, meaning that it will form an arc of a parabola (or at least part of an arc). The reason we state ‘part of an arc’ is that we need to consider the restrictions on x.

First of all x ≥ 0. Then, we are told that the production capacity is 400 units, so we have that x ≤ 400. That is, the profit function has a domain given by 0 ≤ x ≤ 400.

We can then use the turning-point form of the parabola:

From this equation we can see immediately that the maximum occurs at x = 390.

(d) The maximum profit is then $151600.

(e) The unit price is 800 – 390 = 410 dollars.

Note: In this case we have not used the full capacity. If our production limit were 350 units per day then the optimum would have occurred at x = 350 which would not be at the vertex of the arc but at an endpoint.

(a) The manufacturer will break-even when . That is,

[using the quadratic formula]

Therefore, as x is an integer, we can say that at x = 191 and x = 2359, the manufacturer will break-even.

The graph shows the cost function, , and the revenue function, , for a transistor manufacturer, where x is the number of units sold.

(a) Determine the number of transistors that need to be sold for the manufacturer to break-even.

(b) For what values of x will the manufacturer be i. in the black?

ii. in the red?

(c) Determine the maximum profit that the manufacturer can make.

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(b) i. To be in the black means to be making a positive profit –i.e., a gain.

This occurs when or simply, .

From part (a) and the graph, this would occur when 191 ≤ x ≤ 2359.

ii. To be in the red means to be making a negative profit – i.e., a loss.

This occurs when or simply, .

From part (a) and the graph, this would occur when 0 ≤ x ≤ 190 or 2360 ≤ x ≤ 3000.

These results can be shown on the graph below:

(c) The profit equation is given by, .

We determine the maximum value of this quadratic function using the graphics calculator:

We can then say that the maximum profit is $78,375 and occurs when 1275 items are sold.

Note the importance of setting the appropriate window in order to obtain a good visual display of the behaviour of the profit function.

As we have seen in this section, there are times when information, including equations is given.

This information has been obtained by some means and is presented as a summary – which, as we have seen can come in different forms. However, it is then up to you to use this information and from it be able to analyse and provide findings that are consistent with the information presented.

We conclude this section with a quick note on aspects of linear functions that have not been dealt with specifically in previous chapters but have a number of applications in many standard

R x( )>C x( ) P x( )>0

R x( )<C x( ) P x( )<0

0 3000 x 30000

C x( ) = 30000+30x R x( ) x 200 x

15

--- – 

 

=

y (£)

191 2359

Loss

Loss

Gain

Break-even points

P x( ) = R x( )–C x( )

∴P( )x x 200 x 15

--- – 

 –(30000+30x)

= x² 15

---– +170x–30000 0, ≤ ≤x 3000

=

P x( ) x2 15

---– +170x–30000 0, ≤ ≤x 3000

=

1275

Maximum profit 3000

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practices that involve the usage of utilities such as water, electricity and so forth as well as payments for taxi fares, postage and the like.

In document IB Maths SL Textbook (Page 81-88)