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Applied Statistics and Mathematics in Economics & Business

Introduction

BS1501

Functions of more than one variable

Till now we have only considered functions dependent on one variable, e.g. y = 3x + 8. Most

functions involve more than one variable, e.g. profit may depend on labour and time.

A function f of two variables, x and y, can be written f (x, y), e.g. f (x, y) = xy + 12x.

To evaluate this function, we specify the numerical values of the variables, x and y, e.g. if x =

3 and y = 4, then f (3, 4) = 3 4 + 12 3 = 48. Note that f (3, 4) and f (4, 3) probably have

different values, in this case f (4, 3) = 4 3 + 12 4 = 60.

For functions of one variable we constructed differentiation, for functions of more than one

f

variable we use partial differentiation, e.g.

(read as ‘partial dee f by partial dee x’ or

x

sometimes just ‘partial f by partial x’). It is found by differentiating with respect to x, the

associated y is temporarily treated as a constant (acts like a constant). For example,

f (x, y) = 12x + xy then

f

= 12 + y

x

More examples;

f (x, y) = x3 + 36xy

f (x, y) = x3 y + 12xy2

then

then

f

= 3×2 + 36y

x

f

= 3×2 y + 12y2

x

and

and

f

= 36x

y

f

= x3 + 24xy

y

See Exercise 13

Example

Write out all first and second order partial derivatives for the function

f(x, y) = 8×3 y2

We may need to investigate the turning points of these functions of more than variable.

Example

Consider the function f (x, y) = 10x – x2 + 10y – y2 , what can we say about the turning points

of this function? The graph of this function is given below.

From the graph we see that this function

has a maximum (top of the hill shape

shown). Can we find for what values of x

and y is the function at its maximum?

Looking at the arrows labelled 1 and 2

(tangents at the top), we see a point of

intersection where the maximum occurs.

Considering just arrow 1, its slope in the x

direction is 0, similarly for the arrow 2 its

slope in the y direction is 0.

That is, the points of x and y such that the function f (x, y) is optimal is in mathematical terms

when;

f

f

= 0 and

= 0.

x

y

f

f

= 10 – 2x = 0 and

= 10 – 2y = 0, hence there is an optimal ‘turning’

x

y

point at x = 5 and y = 5.

In our example,

Is this optimum a maximum, minimum

or something else, possibly a point of

inflection or saddle point? For example,

f (x, y) = x2 – y2 (graphically shown).

To investigate the nature of the turning points, we need to differentiate the first order partial

derivatives, there are a number of outcomes to this (for f (x, y) = 10x – x2 + 10y – y2 );

2f

= –2,

x2

2f

x2

and

2f

y2

2f

= 0,

x y

are similar to

second order derivatives in one

2f

2f

variable calculus,

and

x y

y x

are the cross partial derivatives,

2f

2f

here, in general

=

.

x y y x

2f

= –2,

y2

2f

= 0,

y x

f

f

= 0 and

= 0, have been satisfied, the following constraints, if met, will be

x

y

sufficient for determining whether an identified turning point is a maximum or a minimum.

Assuming,

2

For a maximum:

2f

2f

2f

2f 2 f

,

<
0,
<
0
and
x2
x2 y 2
y2
x y
For a minimum:
2f
2f
2f 2 f 2f
.
> 0,

> 0 and

x2

x 2 y 2 x y

y2

2

The latter inequality is the associated Discriminant. Hence in our example we have,

2f

2f

=

–2

<
0,
= –2 < 0
x2
y2
2
and
2f
2f
2f
.
=
4
>

0

=

x2

y2

x y

So we can say that at the turning point, x = 5 and y = 5, there is a maximum, which is verified

by the graphical representation of f (x, y).

Example

Find the turning point and its nature for the following function,

f (x, y) = 3×2 + 4xy – 14x – 20y + 4y2 + 1000.

Firstly, partially differentiate with respect to x and y separately, giving (when set to zero),

f

f

= 6x + 4y – 14 = 0 and

= 4x – 20 + 8y = 0.

x

y

Rewriting the above equations, then,

6x + 4y = 14

4x + 8y = 20

Solving these two simultaneous equations we

get the solution, x = 1 and y = 2. That is, there

is a turning point when x = 1 and y = 2.

However, is it a maximum or minimum?

Partially differentiating again we get,

2f

2f

=

6

>

0,

= 8 > 0,

x2

y2

and

2f

= 4,

x y

2

2f

2f

2f

is

and so

x2

y2

x y

satisfied. In this case 6 8 42 so is

also satisfied. It follows, the above

function has a minimum at the point, x

= 1 and y = 2, the value of f (x, y) at this

point is;

f (x, y) = 3 12 + 4 1 2 – 14 1

– 20 2 + 4 22 + 1000

= 3 + 8 – 14 – 40 + 16 + 1000

= 973

See Exercise 14

Applied Statistics and Mathematics in Economics & Business

Lagrange Multipliers

BS1501

Optimisation

Many types of problems or decisions have certain constraints associated with them, in the

form limited capitol, labour, machinery, time or other resources.

The problem of optimising our use of these resources to achieve maximum profit or minimum

labour and time expenditures etc., is of importance to us.

Here we will illustrate one way in which we can maximise (minimise) the values of functions

depending on certain constraints.

Introduction

The method of Lagrange Multipliers aims to optimise an objective function:

f (x, y)

(e.g. f (x, y) = xy)

(throughout this section we will consider objective functions dependent on two variables

only), subject to a constraint:

h (x, y) = M

(e.g. x + 4y = 16)

The graph shows the objective function f (x, y) = xy and constraint x + 4y = 16. In the left

graph, along a curve its value stays the same for different values of x and y (e.g. f (4, 1) = 4

and f (6, 2) = 12).

This family of curves moves in a north east direction as the value of f (x, y) increases. The

straight line represents x + 4y = 16. Our goal is to find the highest valued curve of f (x, y)

which touch’s the straight line (but must otherwise lie on one side of the curve).

In the above example we see that the values x = 8 and y = 2 achieves this, giving the

maximum value of f (x, y) to be f (8, 2) = 16.

How can we do this mathematically, points to note are:

At the constrained optimum point c = (xc, yc), the highest level curve of f (x, y) is tangent to

the constraint h (x, y) = M, i.e. they have the same slope at this point, this can be written as:

f

h

( xc , yc )

(x , y )

x

x c c

=

f

h

( xc , yc )

(x , y )

y

y c c

Rewriting the equation as

where is the common value of the two

quotients. Hence we can rewrite the above

as two equations, i.e.

f

f

( xc , yc )

(x , y )

x

y c c

=

=

h

h

(x , y )

(x , y )

x c c

y c c

f

h

( xc , yc ) −

(x , y ) = 0

x

x c c

f

h

( xc , yc ) −

(x , y ) = 0

y

y c c

there are three unknowns in these two equations i.e. xc, yc and , so we need a third equation.

So we use the equation h (x, y) − M = 0.

Hence we are now in a position to describe the method, which is as follows

i)

Define the new function (called the Lagrangian Function)

g (x, y, ) = f (x, y) + (M − h (x, y))

note: (M − h (x, y)) will not change the value of g (x, y, ), since (M − h (x, y)) = 0.

ii) Solve the simultaneous equations formed by calculating

g

f

h

( x, y , ) =

( x, y ) −

( x, y ) = 0

x

x

x

g

f

h

( x, y , ) =

( x, y ) −

( x, y ) = 0

y

y

y

g

( x, y, ) = M – h (x, y) = 0

to determine the x, y and . The scalar is called the Lagrange Multiplier.

Example

Use Lagrange multipliers to find the optimal value of the objective function

xy

subject to the constraint

x + 4y = 16

Using our theory then we have

f (x, y) = xy,

h (x, y) = x + 4y,

M = 16

the Lagrangian function is given by

g (x, y, ) = xy + (16 − x − 4y)

= xy + 16 − x − 4y

Working out the three partial derivatives of g gives

g

g

g

=y −=0

= x − 4 = 0

= 16 − x − 4y = 0

x

y

hence we need to solve the 3 equations with 3 unknowns

y − =0

x

− 4 = 0

x + 4y

= 16

[1]

[2]

[3]

To find put y = from [1] and put x = 4 from [2], in [3]

4 + 4 = 16

so

8 = 16

giving

= 2.

It follows that y = 2 and x = 8.

Therefore the optimal solution is when x = 8 and y = 2, the corresponding value for the

objective function xy subject to the constraint x + 4y = 16 is f (8, 2) = 8 2 = 16

Up till now we have no idea whether the optimal solution is maximum or minimum, from our

discussion, we have a condition which if satisfied tells us that the function at this point is a

maximum or minimum, this being:

For a maximum:

g

= 0,

x

g

= 0,

y

g

= 0,

2g

< 0,
x2
2g
< 0,
y2
2g 2g 2g
x2 y 2
x y
2
g
= 0,
2g
> 0,

x2

2g

> 0,

y2

2g 2g 2g

x2 y 2

x y

2

For a minimum:

g

= 0,

x

g

= 0,

y

If neither of these sets of conditions hold, then more investigation is needed, this is not

discussed in this course. For this course this is the only check we make to gain information on

whether the optimal solution is a maximum or minimum.

g

g

g

= 0,

= 0,

=0

y

x

since we have an optimal solution when x = 8 and y = 2, so now look at the second

derivatives;

Referring back to our previous example then we already know

2g

2g

=

0,

= 0,

y2

x2

2g

= 1,

x y

so we are unable to determine the nature of the optimal solution using this method.

See Exercise 15

2

2g 2g 2g

hence

8

2 36 +

not scarce

7

7

So the Anthracite and Best coal needs were only just met, calculating their shadow prices;

If we were able to produce one less unit of Anthracite

what would be the increase/decrease in price? First solve

Remove U from equation [2], so: [2] 5 [1]

Giving U =

14L = 5,

, putting these values into the

equation for the cost, C = 100U + 150L, we get:

U + 4L = 7

5U + 6L = 30

hence L =

C = 100

£610.71

[1]

[2]

5

14 hours

+ 150 5

14 =

So decrease in cost would be £621.43 £610.71 = £10.72, hence the Shadow price for

Anthracite is £10.72.

If we were able to produce one less unit of Best coal what

would be the increase/decrease in cost? First solve

Remove U from equation [2], so: [2] 5 [1]

Giving U =

68

equation for the cost, C = 100U + 150L, we get:

[1]

[2]

11 hours

14

11

C = 100 68

14 + 150 14 =

14L = 11,

14 , putting these values into the

U + 4L = 8

5U + 6L = 29

hence L =

£603.57

So decrease in cost would be £621.43 £603.57 = £17.86, hence the Shadow price for Best

coal is £17.86.

See Exercise 31

Applied Statistics and Mathematics in Economics & Business

BS1501

Business applications:

Resource allocation

Linear

Programming

Before we start

Sketch the feasibility area (shading it in, which satisfies) the following inequalities:

y 2,

x + y ≤ 10,

y ≤ 4,

y–x≤1

—————————————-

2. Sketch the feasibility area (shading it in, which satisfies) the following inequalities:

y < 5,
y > 3,

x < 5,
y – 2x > –4

Introduction

Linear programming is a method for finding the optimal solutions to problems of constrained

optimisation.

Many real life problems will mean many variables to consider, here the examples only use 2

or 3 variables. To describe linear programming we consider a simple example.

Example

A farmer has 100 acres of land, and £5,000 capital available, she has the choice of two crops

she can grow on her land, these being wheat and barley. She estimates that wheat costs £40

per acre and barley £60 per acre to seed and the respective profits per acre are £30 (wheat)

and £40 (barley), what mixture of wheat and barley must she grow for maximum profit?

Let W be the number of acres of wheat.

Let B be the number of acres of barley.

Let P be the profit in total.

looking at the land constraint, we see the maximum amount of land

available to us is 100 acres, so our first constraint is that the amount of

land growing wheat (W) or barley (B) cannot exceed 100 acres, i.e.

The amount of money available is £5,000, this amount will be split up

between the number of acres of wheat (£40 per acre) or barley (£60

per acre) which is grown, so our second constraint is:

The amount of profit (P) we achieve is also dependent on the

number of acres of wheat (£30 profit per acre) and barley (£40

profit per acre) grown and is given by:

W + B 100.

40W + 60B

5000.

P = 30W + 40B,

our aim is to maximise

this value P.

We note also that we have two non-negative constraints, these being W 0 and B 0, since

we are unable to grow a negative number of acres of wheat and barley.

Summary table

Land

Cash

Profit

Non-negative

Wheat (W acres)

l (acre per acre)

40 (£ per acre)

30 (£30 per acre)

Barley (B acres)

1 (acre per acre)

60 (£60 per acre)

40 (£40 per acre)

Constraint

W + B 100

40W + 60B 5000

maximise P = 30W + 40B

W 0, B 0

We now turn our attention to constructing a

graphical interpretation to the above constraints:

For the land constraint, when W = 0 then B 100,

similarly when B = 0 then W 100, as shown.

The shaded area is that area of the graph which

satisfies W + B 100 (also satisfies the constraints

W 0 and B 0). In a similar way we could also

draw a graph (and subsequent shaded area) for the

constraint 40W + 60B 5000, see graph.

Combining the two graphs we get a graph of the

form shown.

Where the shaded area in this graph is that area

within which all 4 constraints (inequalities W + B

100, 40W + 60B 5000, W 0 and B 0) are

satisfied (sometimes called the feasible region)

Turning our attention to the profit constraint i.e.

maximise P = 30W + 40B, we can draw a series of

lines for various values of P by writing the previous

equation as B =

3

1

W+

P.

4

40

Any two points on one of these lines has the same

profit value, we see that as the lines move away

from the origin the value of P increases.

The above diagram combines the two previous

graphs, we can find the most profitable point within

the feasible area for P given the by the point A.

This point can be found mathematically by solving

the simultaneous equations

W + B = 100

40W + 60B = 5000

[1]

[2]

From [1], let B = W + 100 into equation [2], giving, 40W + 60(W + 100) = 5000

40W 60W + 6000 = 5000

20W = 5000 6000

20W = 1000

W = 50

which gives B = 50, putting these values into the equation for the profit P = 30W + 40B we

get,

P = 30 50 + 40 50 = £3,500.

What happens if the profit on wheat becomes £40

per acre and the profit on barley becomes £35 per

acre?

The profit function changes to be

P = 40W + 35B,

hence we get a slightly different set of lines denoting

the change in the value of profit

Using this above set of lines and the original feasible

region we have

In this case the optimum point is when

W = 100 and B = 0,

i.e. grow only Wheat. Let us consider another

example.

The change in the solution came about from a change in the gradient of the objective

function, and is an issue on the sensitivity of the solution.

Consider the gradients of the lines which define the

feasible region.

The land constraint B = W + 100, gradient is 1.

For the capital constraint gradient is

2

3

While gradient of objective function line’s gradient is between 1 and

as found. Objective function B =

W+

P, has gradient

, solution will stay

, between these values.

In the change of this example the objective function

became;

P = 40W + 35B,

i.e.

which has a gradient

B=

40

1

W+

P,

35

35

40

, i.e. 1.1428.

35

Hence we would have expected the optimum point

to have changed in this case.

How about at the optimum points, in these cases we have only one bound on the gradient of

2

the optimum function, i.e. at P gradient must be greater than

and at Q gradient must be

3

less than 1.

Example

Jarvis Ltd. makes two kinds of doors ‘Sturdy’ and ‘Deluxe’, but are limited because of

limitations on the amount of, glass, wood and labour, available. The firm can obtain 600m 2 of

wood per week at a cost of £4 per m 2 of wood and 300m2 of glass at a cost of £16 per m 2,

they have 8 operators, each working a 40 hour week, at an hourly rate of £6.

‘Sturdy’ uses 2m2 of wood and 0.5m2 of glass, while ‘Deluxe’ uses 1.5m2 of wood and 1m2 of

glass. It takes 1 hour of labour time to cut and assemble each door. ‘Sturdy’ sells for £30 and

‘Deluxe’ sells for £40.

What is the optimum number of ‘Sturdy’ and ‘Deluxe’ doors to sell that maximises profit?

Summarising data

Wood

Glass

labour

Non-negative

Sturdy (S doors)

2 (m2)

0.5 (m2)

1 (hour)

Deluxe (D doors)

1.5 (m2)

1 (m2)

1 (hour)

Constraint

2S + 1.5D 600

0.5S + 1D 300

S + D 320

S 0, D 0

The wood, glass and labour constraints give us the

following feasibility region

To calculate the profit function we need to remove all the resource costs from the selling

price of each door

Sturdy:

£30 2 £4 0.5 £16 £6 = £8

profit

Deluxe:

£40 1.5 £4 1 £16 £6 = £12 profit

Hence profit (P) is given by;

P = 8S + 12D

In the same way as the previous example we can

create a series of lines for various values of P, any

two points on a single line create the same amount

of profit;

Combining these graphs

We can yet again solve this numerically by first solving where the optimum profit touches

two of lines making up the feasibility region

0.5S + 1D = 300

[1]

1S + 1D = 320

[2]

To remove D,

[2] [1]

0.5S = 20 hence S = 40, so D = 280

Putting these values into the equation for the profit i.e. P = 8S + 12D we get

P = 8 40 + 12 280 = £3,680

Q8.

Q7.

Q9.

zan-hust thε-26 = (hik) f

hчу) = 142

М – ер 50

8 (1, 2) (#2 – зазначу-ч-)

г 8

t» ( 50-х-2)

–

оз – kazg- = 1

ne

.

9.

a

бо

– се

о: 4– a5 =

— 64 + 84

+ 8x -x=0

–

82-84 = 2 о

8 – 4 = 2

e=h-x-es

Now, eq” 0 teq (2)

-)

x=la E r2 =h2

POCO

SMOT ON DOC MODO

94, 95-) – 8

1 – 8 м, 4 )

8y-on-

do

8y-6y=

2lambda

2y=2lmabda

y=lambda

8n-by-d-o

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