Mathematics QM1 : Revision Summary, Revision Exercises and
examination papers for June 2000 and June 2001
NOTE.
Revision summary and exercises
This revision summary is not meant to be exhaustive but a general guide to the
main ideas of `the course. On the last two pages are 15 revision exercises
covering the topics in the summary. Many of them are old examination
questions suitably altered to take account of the different syllabus this year. You
are strongly advised to do these as part of your revision as QM1 is very much a
problem-solving course and practice does indeed make perfect ( solutions will
be available from your tutors from week 1 of the Summer term or available
from 1C4 after week 3). Also the textbook for the course, Bradley and Patton,
offers you a wide selection of worked examples and exercises with answers.
Previous year’s QM1 papers
You are strongly advised to work through June 2000 and June 2001
examination papers (which are attached ) but please note that since June
2000 the syllabus has changed slightly. For example, Chapter 1 (on random
variables and their distributions) was new for the academic year 2000/2001
(in the 1999/2000 academic year it was examined as part of QM2 only),
evaluating second derivatives of implicit functions was removed from the
syllabus so Question 2(ii)(c) on the June 2000 exam paper is not relevant this
year and quadratic Taylor polynomials were removed from the syllabus so that
Question 7(iii) on the June 2000 exam paper is not relevant this year.
Solutions to both QM1 papers will be available from tutors from
week 1 of the Summer term and from 1C4 after week 3.
2
A. Random variables and their (probability) distributions
For a summary see Chapter 1, QM2
B. Linear Algebra
(You would not be expected to prove any results about determinants or matrices)
However , in particular you should be able to
(i)
Add and subtract matrices, multiply a matrix by a scalar, multiply matrices together
where possible i.e. if calculating the product AB, the number of columns of A must
equal the number of rows of B otherwise the matrices are not conformable for
multiplication.
(ii)
Transpose a matrix ( to transpose a matrix interchange the rows and columns - the
transpose of a matrix A is A ).
(iii)
Find the determinant of a matrix (expanding from any row or column) and find the
trace of a matrix.
(iv)
Understand the terms symmetric matrix (i.e. A = A where A is square), diagonal
matrix, identity matrix (denoted I n ), idempotent matrix (i.e. A 2  A ), inverse
matrix A 1 (i.e. AA1  A 1 A  I ), non-singular matrix (i.e. det(A)  0 , A 1 exists)
(v)
Determine whether a symmetric matrix is positive definite (i.e. all the principal
minors are positive) or negative definite (i.e. the principal minors oscillate in sign
with the first principal minor being negative).
(vi)
Invert a 2 x 2 matrix and verify that a particular 3 x 3 matrix is the inverse of a given
3 x 3 matrix .
(vii)
Express a set of equations in 2 or 3 unknowns in the form A x = b .
(viii)
Solve a set of equations using Cramer’s Rule.
(ix)
Solve a set of two equations in 2 unknowns using the inverse matrix method by
finding A 1 and then setting x = A 1 b or solve a set of three equations in 3
unknowns using the inverse method given the inverse matrix A 1 .
Note.
Apart from 2.6.1(b), (d) and (f), 2.6.2 (e) and (f) the results given in section 2.6 are al
given for information only. These particular results will be covered in more detail in certain
second-year modules as and when they are needed. Also there are worked examples and
exercises with solutions to all parts(i) – (ix) above. See Chapter 2 and Exercise 2.
3
C.
Functions
(a)
Homogeneous functions
degree r
if
A function f = f ( x1 , x 2 ,..., x n ) is homogeneous of
f  x1 ,  x2 ......,  xn   r f ( x1 , x 2 ,..., x n ) .
Also there is a connection with economies of scale and the value of r.
e.g. If f is a production function, then if r < 1, you have decreasing returns to scale,
if r = 1 you have constant returns to scale and if r > 1 you have increasing returns to
scale.
Euler’s theorem
If f is homogeneous of degree r then
f
f
f
.x1 
.x 2  ............. 
.x n = r f ( x1 , x 2 ,..., x n ) i.e. r f
x1
x 2
x n
In the case of n = 2 i.e. f = f ( x1 , x 2 ) this becomes
f
f
.x1 
. x 2 = r f ( x1 , x 2 )
x1
x 2
(b)
Monotonic functions These are functions which are strictly increasing, increasing,
strictly decreasing or decreasing functions.
e.g. Suppose f = f (x) , then f is increasing if for all real values x and x 
f ( x)  f ( x  ) for all x  x 
and strictly increasing if f ( x)  f ( x  ) for all x  x 
(c)
Linear functions
A function f = f ( x1 , x 2 ,..., x n ) is linear if it can be written
n
f ( x1 , x2 ,..., x n )  a1 x1  a2 x2 ... an x n   ai xi with x 0 =1.
i0
For the case n =1, i.e. f = f (x) this becomes f (x) = a0  a1 x
(d)
Implicit functions
The equation f x1 ,x2  = c  may define x 2 as an implicit function of x1 . Suppose
this implicit function of x1 is denoted by g x1  so that x 2  g x1  .
f
dx 2
 x1  f 1
 g ' ( x1 ) 

f
dx1
f2
x2

Then
............ eqn. (*)
4
This result is obtained by using the chain rule.
You should know how to prove this result as well as use it.
(d)
Continuous functions of one variable
If the function is continuous there are no gaps in the graph of it.
(e)
Concave and convex functions
Concave and convex functions are particularly important in constrained optimisation
as it is assumed that the objective function and the constraints are concave so that the
Lagrangean is also concave.
Any stationary point of a differentiable concave (convex) is a global maximum
(minimum) so we are effectively finding a stationary point of the Lagrangean
function in optimisation.
A function f  x1 , x2 ,....., xn  is concave if the Hessian matrix H is negative definite
and convex if H is positive definite where H is the (n x n) symmetric matrix of
second-order partial derivatives of f .e.g.
 f11
H 
 f12
 f11
f12 

if n = 2 or H  f12


f 22 
 f13
f12
f 22
f 23
f13 
f 23  if n = 3.
f 33 
You should be able to show that a function of 2 or 3 variables is concave or convex .
(see Exercise 5, No.10, Examples 5.10 and 5.16 in Chapter 5 in lecture notes and
Revision Exercises, No. 12(i)).
B.
Contours
Suppose we have a function f = f  x1 , x2  , then a typical contour
of f  x1 , x2  is f  c  where c  is some number (e.g. 2 , see Revision exercises
No. 3).
A contour is a curve on which the value of f is constant and equal to c  . As you
change x1 and x 2 but keep c  the same, you move around the contour f  c  .
At all points on the contour the value of f is c  . If c  is increased so is the value
of f and as c  is decreased so is the value of f. Also as c  changes you get different
contours of f.
The equation of a contour is also given by x 2  g x1  (see section A(c) above) .
5
Economic examples of contours
(i)
(ii)
(iii)
(iv)
indifference curves where f = a utility function,
isoquants where f = a production function ,
isocost curves where f = a cost function and
isoprofit curves where f = a profit function.
You should be able to sketch a contour - see Bradley and Patton, Chapter 7 as well
as lecture notes and Exercise 3.
Interpretation of equation (*) The equation f  x1 , x2  = c  defines a contour of f
and the equation of the contour is x 2  g x1  so that (*) is the slope of the
contour at the point  x1 , x 2  .
If f = a utility function u x1 , x 2  then the slope of an indifference curve (i.e. a
contour of f ) is - M.R.S.
Note that M.R.S. is the marginal rate of substitution and is defined as
u
 x1 u1

M.R.S. =
.
 u u2
 x2
If f = a production function Q  f l , k  then the slope of an isoquant
is - M.R.T.S.
Note that M.R.T.S. is the marginal rate of technical substitution and is defined as
Q
f
Q
M.R.T.S. = l = l = l
Q Q k f
k
k
6
E
Differentiation
(a)
One variable Suppose f = f (x) and x and x are any two real numbers.
f  x  f  x 
f
or lim
exists we say that f is differentiable at x .

x

0
x
xx
We call the limit the derivative of f at x . If f is differentiable at all x then we say
If lim
x x
f that is differentiable and we denote the derivative by f (x).
If f is a differentiable function then it is continuous but not vice-versa i.e. not all
continuous functions are differentiable. It follows that if f is discontinuous it is also
not differentiable.
Derivatives of standard functions
d
(xn) = n xn-1 for all values of n`
dx
d g x 
e
 g   x  e g x 
(ii)
dx
d
(ax) = ax ln a
dx
f   x
d
(iv)
(ln f  x  )) =
f  x
dx
2
d f
We denote the second derivative of f by f (x) or
.
dx 2
d n f  x
Similarly, the nth derivative of f is denoted f (n)  x  or
.
dx n
(i)
(iii)
Applications to economic functions
(i)
Price elasticity of demand  d 
d ln Qd
dQd p
or
where Q d is the
.
d ln p
dp Qd
quantity demanded.
(ii)
Price elasticity of supply
s 
dQs p
.
dp Qs
or
d ln Qs
where Qs is the
d ln p
quantity supplied.
(iii)
(iv)
Marginal utility MU = u  x  
du
where x is the quantity consumed .
dx
dC
Marginal cost MC = C  x  
where x is the level of output and C is the
dx
total cost function .
(v)
Marginal propensity to consume = MPC = C Y  
dC
where C is the
dY
consumption function and Y is the level of national income.
(vi)
Marginal propensity to save = MPS = S Y  
dS
where S is the saving
dY
function
(S(Y) = Y - C(Y) ) and Y is the level of national income.
7
(vii)
Marginal product = MP =
dQ
where Q is a short-run production function which
dl
is a function of the labour input, l.
(viii)
Marginal revenue = MR = R Q  
dR
where the revenue R is a function of Q,
dQ
the output.
(ix)
Marginal revenue product = MRP = R l  
dR
where revenue R is a function of l
dl
the labour input
Connection between MRP, MR and MP
MRP = MR . MP
Points of inflection
Given a function f (x), then the point x  will be
(i) a stationary point of inflection if
 
 
 
f  x   0 , f  x   0 and f  x   0
 
(ii) a non-stationary point of inflection if f  x   0
,
 
 
f  x   0 and f  x   0 .
 
Note that a stationary point is a point x  where f  x   0
(b) Several variables Suppose that f = f ( x1 , x 2 ,..., x n ) and we require the partial
derivative of f with respect to x i , denoted by
f
or f i . We differentiate f
 xi
allowing x i to change but keeping all the other (n -1) variables constant.
Example. Suppose n = 3 and f = f ( x1 , x 2 , x 3 ) then there are 3 first-order partial
derivatives,
f
f
f
or f 1 ,
or f 2 ,
or f 3 .
 x2
 x3
 x1
There are 9 second-order partial derivatives, namely
2 f
or f 11 ,
 x12
2 f
or f 22 ,
 x22
2 f
or f 12 ,
 x1 x2
2 f
or f 13 ,
 x1 x3
2 f
or f 21 ,
 x2 x1
2 f
or f 31 ,
 x3 x1
2 f
or f 33 ,
 x32
2 f
or f 23 ,
 x2 x3
2 f
or f 32 .
 x3 x2
The last 6 are known as cross-partial derivatives or cross-partials and
f 12 = f 21 , f13 = f 31 and f 23 = f 32 .
8
Interpretation. Suppose n = 2 and f = f  x1 , x2  then
f
is the gradient of the surface given by f in the x1 direction.
 x1
f
is the gradient of the surface given by f in the x 2 direction.
 x2
Suppose that f = f ( x1 , x 2 ,..., x n ) and we now allow all the n variables to change,
we arrive at the idea of a total differential.
Total differential (the formula (*) will be given in exam if needed)
Suppose f is a function of n variables i.e. f = f  x1 , x 2 ,....., x n  .
The total differential of f at the pt. ( x1 , x 2 ,..., x n ) is denoted by df  x1 , x 2 ,....., x n 
and is defined by
df  x1 , x 2 ,....., x n   f1 ( x1  x1 )  f 2 ( x2  x2 ) 
 f n ( xn  xn )
(*)
where f1 , f2 ,....., fn (the n first-order partial derivatives) are all evaluated at the point
( x1 , x 2 , ..... x n ).
It represents the approximate change in f as we move from the point ( x1 , x 2 , ...
xn )
to any point  x 1 , x 2 ,......, x n  .
It represents the approximate change in f as we approximate f by some form of linear
approximation. It will only be a good approximation close to ( x1 , x 2 , ... x n ) .
N.B. 1. The function f has n first-order partial derivatives but only 1 total differential.
2. The partial derivatives must be evaluated at the point ( x1 , x 2 , ...
xn ) .
Special cases of interest
df ( x )  f ( x )( x  x ) = differential at x  x .
n=1
f = f (x) ,
n=2
f = f  x1 , x2  , df ( x1 , x 2 ) 
f
f
( x1  x1 ) 
(x  x2 )
x1
x 2 2
= total differential at ( x1 , x 2 )
Important notes
(i) The total derivative is not the same as the total differential.
e.g. Suppose f  f x1 , x2  and x2  g x1  , then the total derivative with respect to
x1 at the point x1 , x2  is
df
evaluated at the point, which will give a single value,
dx1
whereas the total differential at the point x1 , x2  is df x1 , x2  (see above) and is a
linear equation in x1 and x 2 .
9
(ii) The total derivative is not the same as the partial derivative unless n = 1 i
.e. f  f x  .
e.g. Suppose f  f x1 , x2  then
df
f

.
dx1 x1
With total derivatives we allow all the variables to change (e.g . both x1 and x 2 )
whereas with the partial derivative we allow only one variable to change (e.g . x1 )
Application to economic functions
(i) The own (or direct ) price elasticity of demand
 d or  1 
 Q1 p1
 x1 p1
 ln Q1
 ln x1
.
.
=
or  1 
=
 p1 Q1
 p1 x1
 ln p1
 ln p1
(ii) The cross price elasticity of demand
 c or  12 
 Q1 p2
 x1 p2
 ln Q1  ln x1
.
.
=
or 12 
=
 p2 Q1
 p2 x1
 ln p 2  ln p 2
Notice that if  c or  12 < 0, then goods 1 and 2 are complementary goods and if
 c or  12 > 0, then goods 1 and 2 are substitute goods. Also if the  c or  12 is
close to zero, then the goods 1 and 2 bear little relationship to each other.
(iii)
The income elasticity of demand
Y 
 Q1 Y1
 x1 Y
 ln Q1  ln x1
.
.
=
or  Y 
=
 Y Q1
 Y x1
 ln Y
 ln Y
In all these definitions Q1 or x1 is the quantity demanded of good 1, p1 and p 2 are
the prices of goods 1 and 2 respectively and Y is the consumer’s income.
(For examples see Exercise 5, Nos. 8 and 9 and Revision exercises Nos. 7 and 13)
(iv)
Partial elasticity of output with respect to capital input =  Qk 
 ln Q
Q k
.
. or
 ln k
k Q
(v)
Partial elasticity of output with respect to labour input =  Ql 
 ln Q
Q l
.
. or
 ln l
l Q
where k is the capital input, l is the labour input and Q is the production function which
10
is a function of l and k e.g. Q  Al  k  .
(For examples see Exercise 5, 2(ii) and Revision exercises No. 5)
(vi)
(vii)
Q
.
l
Q
The marginal product of capital (MPK) =
.
k
The marginal product of labour (MPL) =
Stationary point for a function of n variables
A stationary point of a function f  x1 , x 2 ,....., x n  is any ( x1 , x 2 ,..., x n ) such that
f i ( x1 , x 2 ,..., x n ) = 0 for i = 1,2,......n.
Unconstrained optimisation


Given a function f  x1 , x 2  the necessary and sufficient conditions for a point x1 , x 2 to be


a global minimum are that x1 , x 2 is a stationary point and this stationary point is a
minimum i.e.
 f11 f12 
f1  0 , f 2  0 and the Hessian matrix H  
 is positive definite.
 f12 f 22 
or
f1  0 , f 2  0 and
f11  0 , f11 f22  f122 > 0


Given a function f  x1 , x 2  the necessary and sufficient conditions for a point x1 , x 2 to be


a global maximum are that x1 , x 2 is a stationary point and this stationary point is a
maximum i.e.
 f11 f12 
f1  0 , f 2  0 and the Hessian matrix H  
 is negative definite.
 f12 f 22 
or
f1  0 , f 2  0 and
Conditions for a saddle point
f11  0 , f11 f22  f122 > 0
f1  0 , f 2  0 , and f11 f22  f122 < 0
You should be able to solve economic problems involving maximisation and minimisation
where there are functions of one or two variables and no constraints (see Examples 4.13,
4.14, 4.15, 5.16 - 5.19 in lecture notes and also Revision exercises 10 and 11).
11
F. Constrained optimisation
The whole of chapter 6 is important so I will not summarise it. However there are
several important notes.
(i)
Equality constraints are always binding so you do not have to check that you
have
reached the optimum. Binding means that the constraint has an effect on the
optimisation and if the constraint were not there, the optimum would be
different.
(ii)
With inequality constraints, you must check that you have reached the optimum
by calculating the value of the Lagrange multiplier at the optimum i.e.   . If
  0 you have reached the optimum. If  < 0, then you have not and you
must return to the Lagrangean and put  = 0 i.e. the problem becomes an
unconstrained optimisation.
(iii)
If   = 0 this means the constraint is not binding whereas if   > 0 it means the
constraint is binding.
(iv)
Always state your objective function (the function you are trying to maximise)
and constraint (in the correct form) at the beginning of any solution.
(v)
You always maximise, so if you want to minimise C say, you maximise -C.
(vi)
Generally the Lagrange multiplier at the optimum,   is equal to the change in
the maximum value of the objective function for a marginal relaxation in the
constraint.
(vii)
If your objective function is a utility function (e.g. ux1 , x2   Ax1 x2 ) and the
constraint is the usual linear budget constraint (e.g. p1 x1  p2 x2  m ) ,   is
the
marginal utility of money when the utility is maximised. Also at the optimum,
the M.R.S = the price ratio of the two inputs =
p1
where p1 and p 2 are the
p2
prices of goods 1 and 2 respectively.
(ix)
If your objective function is a negative cost function (e.g.  C  wl  rk ) and
the constraint is a non-linear constraint on output e.g. Q  f l , k   Al  k   q 
so that you require to minimise cost subject to a minimum output,   is the
marginal cost of production when the cost is minimised. Also at the optimum,
the M.R.T.S. = the price ratio of the two inputs =
w
where w is the wage rate
r
(price of labour input) and r is the rental price of capital (price of capital input).
12
C.Osborne March 2002