Worked examples
33
(0.9S)Yt1. During year t, N tonnes of new grain are added, and so the size
of the grain mountain at the end of t years is given by
Yt
=
(0.9S)Ytl
+ N.
We are given that there are initially 30000 tonnes, so Yo = 30000.
Thus we have a r
27. Firstorder differential equations
27.1
Continuoustime models
In Chapter 23 we looked at the dynamics of the economy, using what are
known as discretetime models. This means that the timeperiods involved
were taken to be successive calendar years (
Worked examples
or
=
24
J
353
16c
Now, p(O) = 6 and so when t = 0, 24  (8/3)p = 8. This shows that we must
take the positive squareroot, and that 8 =: J3/16c. Therefore, 16c = 3/64
and
8
24 p =
3
16t + 3/64
3
Rearranging, we obtain the following for
25. Areas and integrals
25.1
The consumer surplus
A typical downwardsloping demand set D is illustrated in Figure 25.1. As
the number of units of the good increases, the price consumers are prepared
to pay for each unit decreases.
p
D
q
Figure 25.1: A ty
Functions
13
Example Let X, Y,Z denote the sets defined above, that is
X=cfw_2,3,5,7,8,9,
Y=cfw_1,4,5,7,9,
Z=cfw_1,2,3,4.
Then we have, for example
Xu Y = cfw_1,2,3,4, 5,7, 8,9,
Y uZ = cfw_1,2,3,4,5, 7,9,
X n Y = cfw_5,7,9,
X n (Y nZ)
We also have relatio
Integration by parts
Example Calculate
1
5
333
x+ 1
+ 2x + 5
2   d x .
o x
The difficulty here arises from the complicated denominator. So, a possible
substitution is x 2 + 2x + 5 = t, which implies (2x + 2) dx = dt. For the indefinite
integral, we ha
Business cycles
24.3
303
Business cycles
We now return to the dynamics of the simplified national economy, as
described by the 'multiplieraccelerator' equations in Section 23.2. As we shall
see, the solution to this model can exhibit oscillatory behaviou
Worked examples
293
Worked examples
Example 23.1 Find the general solution of the recurrence equation
Yt  6Ytl
+ 5Yt2 = 0.
Solution: The auxiliary equation is z2  6z + 5 = 0, that is (z  5)(z  1)
with solutions 1 and 5. The general solution is there
Exercises
283
Exercises
A consumer purchases quantities of two commodities, fruit
and chocolate, each month. The consumer's utility function is
Exercise 22.1
for a bundle (Xl, X2) of Xl units of fruit and X2 units of chocolate. The
consumer has a total of
Nonhomogeneous equations
363
Example Find the general solution of the equation
d2 y
dt 2

dy
4 dt
Find also the solution for which y(O)
+ 13y = 0.
= 1 and y' (0) = o.
The auxiliary equation is z2  4z + 13 = 0, which has no solutions. We have
= 4/2 = 2
Exercises
373
Exercises
Exercise 28.1 Find the general solutions of the following equations.
(b)
d2 y
dt 2

+ 4y =
O.
(c)
d2 y
dt 2
+ 2 dt + 2y =
O.
dy
4 dt
dy
Exercise 28.2 Find a particular solution, in the form y
=
(At + B)e 3t , of the
equation
d2 y
Exercises
53
Main topics
the cobweb model in general terms
how to derive a recurrence equation when demand and supply are linear
solving this recurrence to find the sequences of prices and quantities
analysing the stability of the cobweb model
Key ter
The logarithm function
73
General properties of the logarithm function can be deduced from the corresponding properties of the exponential function. For example, from the rule
exp(x) exp(y) = exp(x + y) it is easy to deduce (Example 7.5) that
In(ab)
=
In
Worked examples
Example 6.2
function f(x)
=
63
Use the derivative to find the approximate change in the
x 4 when x changes from 3 to 3.005. Compare this with the
actual change.
Solution: The derivative of f is f'(x) = 4x 3 and we therefore have the
approx
Worked examples
43
Example 4.3 An amount of $1000 is invested and attracts interest at a rate
equivalent to 10% per annum. Find the total after one year if the interest is
compounded (a) annually, (b) quarterly, (c) monthly, (d) daily. (Assume the
year is
Main topics/Key terms
23
Rearranging,
q2
+ 2q + 27 
75
+ 5q2 + 10q = 0,
that is 6q2 + 12q  48 = 0.
Dividing by 6, we get q2 + 2q  8 = 0, which factorises as (q + 4)(q  2) = O.
The solutions are q = 4 and q = 2 and the corresponding values of p, given
Answers to selected exercises
21.3 The cost function is C(q)
500/
approximately 288.675.
/3,
21.4 S
= 2/3q2. The optimal production level is
L U cfw_(L,p) I p> 5J6L3/ 2 .
= cfw_(q,5J6q3/2) I q
21.5 S = cfw_(O,p) I p
383
Z U cfw_(L,p) I p > Z.
21.6 3125(
Worked examples
323
Worked examples
Example 25.1 Find the area enclosed by the lines t
the graph of the function f(t) = et .
=
1, t
= 2, the taxis, and
2
Solution: The required area is equal to the definite integral A = J1 et dt. Now,
J et dt is one of t
The price ratio and the tangency condition
273
Eliminating A we get
Pi
Pj 
OXi
OXj
This says that the price ratio Pi/Pj is equal to another ratio, known to
economists as the marginal rate of substitution, evaluated at the optimal point.
In the case n = 2
Worked examples
263
p
s
ZL!l
L
L
q
q
(b)
Figure 21.4: the case of decreasing returns to scale
Worked examples
Example 21.1 A firm's weekly output is given by the production function
q(k,l) = k 3/ 4 11/ 4, and the unit costs for capital and labour are v =
21. Constrained optimisation
21.1
The elementary theory of the firm
You will recall that a firm can be described in terms of a function whose
inputs are amounts k (capital) and I (labour), and whose output is a quantity
q (production). The value of q depe
Exercises
183
Find recurrence equations for an, bn, Cn, dn, and solve them to determine an
explicit formula for An.
Exercise 15.5 Find a general formula for B n, where
B=
(11)
0
Exercise 15.6 Suppose that the matrix
of Section 15.3 is
1.05
R=
1.05
(
1.37
Exercises
123
Exercises
Exercise 11.1 Find the first and second partial derivatives of the following
functions:
y2
Exercise 11.2 If f(x,y) = x
note that x y2 = exp (y2 In x) .
,
find % and 0,. [Hint: to help calculate
*,
Exercise 11.3 Suppose that f(x,y)
Choosing optimal bundles
163
In order to see what this means, consider Figure 14.4. Here a and bare
bundles on the same indifference curve u(x) = c, and the assumption requires
that every point on the straight line segment joining them is in Uc . This
mea
Worked examples
and
dl
dk
!k I / 211/ 2
lk l / 2 1 1/ 2
2
The gradient at (1, 1) is, therefore 1/1
=
I
k
1.
Similarly, the derivative dl / dk on an isoquant v(k, I)
dl
dk
1/1
 k/12
VI
V2
133
= c is
k'
o
which is 1 at (1,1).
Example 12.4 The notatio
11. Partial derivatives
11.1
Functions of several variables
Recall that a function f may be thought of as a 'black box', which accepts
an input x and produces an output f(x). In this chapter we shall look at
functions for which the input consists of a pai
Critical points
83
y
a
b
d
x
Figure 8.2: A function with four critical points
We can decide the nature of a given critical point by considering what happens
to f' in its vicinity. Looking at the point a in Figure 8.2 we observe that the
gradient (the deri
Main topics/Key terms
Main topics
revenue and profit
finding critical points
classifying critical points using second derivative
optimisation in an interval
Key terms, notations and formulae
revenue, R(q) = qP(q)
profit function, I1(q)
critical poi
Exercises
103
Exercises
Exercise 9.1 Calculate the elasticity of demand when the demand function is
given by
qD(p) = 70  4p.
For what range of values ofp is your expression valid, and for which of these
values is the demand inelastic?
Exercise 9.2 Show t