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8 (a) 16
(b) Presented with the calculation, − 42 , your
calculator uses BIDMAS, so squares first to get
16 and then subtracts from zero to give a final
answer, − 16. To obtain the correct answer you
need to use brackets:
( 2 4 ) x^2
9 (a) 9 (b) 21; no 10 (a) 43.96 (b) 1.13 (c) 10.34 (d) 0. (e) 27.38 (f) 3.72 (g) 62.70 (h) 2. 11 (a) 7 x − 7 y (b) 15 x − 6 y (c) 4 x + 12 (d) 21 x − 7 (e) 3 x + 3 y + 3 z (f) 3 x^2 − 4 x (g) − 2 x − 5 y + 4 z 12 (a) 5(5 c + 6) (b) 9( x − 2) (c) x ( x + 2) (d) 4(4 x − 3 y ) (e) 2 x (2 x − 3 y ) (f) 5(2 d − 3 e + 10) 13 (a) x^2 + 7 x + 10 (b) a^2 + 3 a − 4 (c) d^2 − 5 d − 24 (d) 6 s^2 + 23 s + 21 (e) 2 y^2 + 5 y + 3 (f) 10 t^2 − 31 t − 14 (g) 9 n^2 − 4 (h) a^2 − 2 ab + b^2 14 (a) 6 x + 2 y (b) 11 x^2 − 3 x − 3 (c) 14 xy + 2 x (d) 6 xyz + 2 xy (e) 10 a − 2 b (f) 17 x + 22 y (g) 11 − 3 p (h) x^2 + 10 x 15 (a) ( x + 2)( x − 2) (b) ( Q + 7)( Q − 7) (c) ( x + y )( x − y ) (d) (3 x + 10 y )(3 x − 10 y ) 16 (a) 4 x^2 + 8 x − 2 (b) − 13 x
17 S = 1.2 N + 3000 E + 1000( A − 21); $204 000
18 (a) C = 80 + 60 L + K (b) C = 10 + 1.25 x (c) H = 5 a + 10 b (d) X = Cd + cm
Exercise 1.1*
1 (a) 3 (b) 5 (c) − 7 2 (a) 2 − 7 − (9 + 3) = − 17 (b) 8 − (2 + 3) − 4 = − 1 (c) 7 − (2 − 6 + 10) = 1 3 (a) − 6 (b) 6 (c) − 5 (d) − 96
(e) − 1 (f) 6 (g) 5 4
(h) 63
4 (a) 6 (b) 2 (c) 5 5 − y^2 + xy − 5 x + 2 y − 6
ChApter 1
Section 1.
Practice Problems
1 (a) − 30 (b) 2 (c) − 5 (d) 5 (e) 36 (f) − 1 2 (a) − 1 (b) − 7 (c) 5
(d) 0 (e) − 91 (f) − 5 3 (a) 19 (b) 1500 (c) 32 (d) 35
4 (a) x + 9 y (b) 2 y + 4 z (c) not possible
(d) 8 r^2 + s + rs − 3 s^2 (e) − 4 f
(f) not possible (g) 0
5 (a) 5 z − 2 z^2 (b) − 3 y (c) z − x^2 6 (a) 7( d + 3) (b) 4(4 w − 5 q )
(c) 3(2 x − y + 3 z ) (d) 5 Q (1 − 2 Q ) 7 (a) x^2 + x − 6 (b) x^2 − y^2 (c) x^2 + 2 xy + y^2 (d) 5 x^2 − 3 xy + 5 x − 2 y^2 + 2 y 8 (a) ( x + 8) ( x − 8)
(b) (2 x + 9) (2 x − 9)
Exercise 1.
1 (a) − 20 (b) 3 (c) − 4 (d) 1
(e) − 12 (f) 50 (g) − 5 (h) 3 (i) 30 ( j) 4 2 (a) − 1 (b) − 3 (c) − 11 (d) 16
(e) − 1 (f) − 13 (g) 11 (h) 0 (i) − 31 ( j) − 2
3 (a) − 3 (b) 2 (c) 18 (d) − 15 (e) − 41 (f) − 3 (g) 18 (h) − 6 (i) − 25 ( j) − 6
4 (a) 2 PQ (b) 8 I (c) 3 xy (d) 4 qwz (e) b^2 (f) 3 k^2
5 (a) 19 w (b) 4 x − 7 y (c) 9 a + 2 b − 2 c (d) x^2 + 2 x (e) 4 c − 3 cd (f) 2 st + s^2 + t^2 + 9 6 (a) 10 (b) 18 (c) 2000
(d) 96 (e) 70 7 (a) 1 (b) 5 (c) − 6 (d) − 6
(e) − 30 (f) 44
6 (a) 2 x − 2 y (b) 2 x (c) − 2 x + 3 y
7 (a) x^2 − 2 x − 24 (b) 6 x^2 − 29 x + 35
(c) 6 x^2 + 2 xy − 4 x (d) 12 − 2 g + 3 h − 2 g^2 + gh (e) 2 x − 2 x^2 − 3 xy + y − y^2 (f) a^2 − b^2 − c^2 − 2 bc
8 (a) 3(3 x − 4 y ) (b) x ( x − 6)
(c) 5 x (2 y + 3 x ) (d) 3 xy ( y − 2 x + 4) (e) x^2 ( x − 2) (f) 5 xy^3 (12 x^3 y^3 − 3 xy + 4)
9 (a) ( p + 5)( p − 5) (b) (3 c + 8)(3 c − 8)
(c) 2(4 v + 5 d )(4 v − 5 d ) (d) (4 x^2 + y^2 )(2 x + y )(2 x − y )
10 (a) 112 600 000 (b) 1.
(c) 283 400 (d) 246 913 577
11 (a) π = 12 y − 3 x − 800 (b) $
(c) 0 ≤ y ≤ x (d) π = 9 x − 800
12 (a) 2 KL ( L + 2) (b) ( L − 0.2 K )( L + 0.2 K )
(c) ( K + L ) 2
Section 1.
Practice Problems
1 (a) 3 5
(b) 4 5
(c) 1 2 y
(d) 1 2 1 3 x
(e) 1 x 2 4
2 (a) 3 8
(b) 7 4
(c) 3 4
(d) 1 18
3 (a)
7 (b)^
15 (c)^
4 (a) 5 x 1 2
(b) x ( x^^1 1) x 1 10
(c) 5 x 1 1
(d) (^) ( x 1^ x 1)(^^1 x^3 1 2)
5 (a) 6 (b) 12 (c)
2 4 (d)
5 (e)^
6 (a) 12 > 9 (true) (b) 12 > 6 (true)
(c) 3 > 0 (true) (d) same as (c)
(e) 2 > 1 (true) (f) − 24 > − 12 (false)
(g) − 6 > − 3 (false) (h) − 2 > − 1 (false)
(i) − 4 > − 7 (true)
7 (a) x > − 7 (b) x ≥ 2
Exercise 1.
1 (a) 1 2
(b) 3 4
(c) 3 5
(d) 1 3
(e) 4 3
2 (a)
20 ;^
25 (b)^
3 (a)
x (b)
2 x (c)^
ac (d)^
3 xy (e)^
a b
4 (a) p 2 q 1 3 r
(b) 1 x 2 4
(c) b 2 a 1 1
(d) 2 3 2 e
(e) 1 x 2 2 5^1 2 x 2 2 2( x 2 1) 2
x 2 1 5 x 2 15 ; other two have no common
factors on top and bottom.
6 (a) 3 7
(b) 1 (^2 ) (c) 5 6
(d) 7 20
(e) 7 18
(f) 5 6
(g) 5 8
(h) 2 5
(i) 7 12
( j) 1 30
(k) 2 27
(l) 21 2
8 (a) 1 x
(b) 2 5
(c) 3 x^^2 x^2
(d) xy
7 y 1 2 x (^) (e) 3 (f) 15 c 1 10 d 36
(g) x^^1 x 1 3
(h) 18 7
h^2 (i) t 20
( j) 1
9 (a) 5 (b) 6 (c) 18 (d) 2
(e) 10 (f) − 1 (g) 60 (h) 1 2 3
(i) − 5 ( j) − 3 (k) − 2 (l) 2 3
(m) 3 1 4
(n) 3 (o) 1 4
10 (a), (d), (e), (f)
11 (a) x > 1 (b) x ≤ 3 (c) x < − 3 (d) x > 2
12^23 x
13 (a) 7 26
2 (b) x ≤ 10
Exercise 1.2*
1 (a) x^^2 2
(b) 2 x 2 1
(^3) (c) 4
(d) − 1 (e) 1 x 2 6
(f) x^^1 x 1 4 (g) 1 2 x^2 2 5 x 1 3
(h) 3
2 x 1 5 y
2 (a) 5 7
(b) 1 10
(c) 3 2
(d) 5 48
(e) 22 13
(f) 11 9
(g) 141 35
(h) 34 5
(i) 6 ( j) 17 10
(k) 7 9
(l) 4
(2, 0) 1
1
2
3
4
5
2 3 4 5 (0, 2 1)
x 2 2 y 5 2
x
y
21 21 22 23 24 25
25242322
Figure S1.
5 From Figure S1.5, the point of intersection is (1, −^1 / 2 ).
(^11)
2
3
4
5
2 3 4 5 x
y
21 21 22 23 24 25
25242322 point of intersection (1, 2 12 )
3 x 2 2 y 5 4
x 2 2 y 5 2
Figure S1.
6 (a) a = 1, b = 2. The graph is sketched in Figure S1.6.
(b) a = −2, b = 1 / 2. The graph is sketched in Figure S1.7.
1
1
2
3
2 3 x
y 1 unit intercept 1 unit
y 5 x 1 2
21 21 22
23 22
Figure S1.
x
y
21
21
22
23
2 1
1
1
y 5 122 2 x
1 2
(^2112)
(^22 )
1 2
1 2 (^2 )
1 2
(^1 )
1 unit
2 units
intercept
Figure S1.
Exercise 1.
1 From Figure S1.8, the point of intersection is (2, 3).
y
x
(2, 3)
Q (^) R
S
P 1
1
2
3
4
5
6
7
8
9
21 21 2 3 4 5
22
25242322
Figure S1.
2 (a) The graph is sketched in Figure S1.9.
Cost in $
1000 900 800 700 600 500 400 300 200 100 1000 2000 3000 4000 5000 Distance in km
Figure S1.
(b) (i) $540 (ii) 2500 km 3 A, C, D, E 4 (a) 6 (b) −1; (6, 2), (1, −1)
5 x y
The graph is sketched in Figure S1.10.
1
1
2
3
4
5
6
7
8
9
2 3 4 5 6 7
4 x 1 3 y 5 24
x
y
Figure S1.
6 (a) (−2, −2) (b) (2, 1^1 / 2 ) (c) (1^1 / 2 , 1) (d) (10, −9)
7 (a) 5, 9 (b) 3, − 1 (c) −1, 13 (d) 1, 4
(e) 5 2 2, (^2) (f) 5, − 6
8 (a) The graph is sketched in Figure S1.11.
(b) The graph is sketched in Figure S1.12.
1 unit 1 unit
1
1
2
3
2 3
y 5 2 x
x
y
21 21 22 23
23 22
Figure S1.
1 unit 2 units
1 1 2 3 4 5 6 7
2
3
4
5
x
y
(^252423222211) 22 23 24 25
y 5 12 x 2 3
26
Figure S1.
9 (a) C = 4 + 2.5 x
(b) The graph is sketched in Figure S1.13.
10
10 15 20 Distance in miles
5
20
30
40
50
Cost in $
Figure S1.
(c) 8 miles
Section 1.
Practice Problems
1 (a) 0 (b) 48 (c) 16 (d) 25 (e) 1 (f) 17
The function g reverses the effect of f and takes you
back to where you started. For example, when 25 is
put into the function f , the outgoing number is 0; and
when 0 is put into g , the original number, 25, is
produced. We describe this by saying that g is the
inverse of f (and vice versa).
2 The demand curve that passes through (0, 75) and
(25, 0) is sketched in Figure S1.16. From this diagram
we see that
(a) P = 6 when Q = 23
(b) Q = 19 when P = 18
75
18
19 23 25
6
P
Q
P 5 2 3 Q 1 75
Figure S1.
3 (a) Q = 21 and P = 36
(b) Q = 18 and P = 48. The consumer pays an
additional $12. The remaining $1 of the tax is
paid by the firm.
4 P 1 = 4, P 2 = 7, Q 1 = 13 , Q 2 = 14.
The goods are complementary.
Exercise 1.
1 (a) 21 (b) 45 (c) 15 (d) 2
(e) 10 (f) 0 ; inverse
2 The supply curve is sketched in Figure S1.17.
(a) 11 (b) 9
(c) 0; once the price falls below 7, the firm does not
plan to produce any goods.
P 5 13 Q 1 7
11 10
7
9 12
P
Q
Figure S1.
3 (a) Demand is 173. Additional advertising
expenditure is 12.
(b) Normal
4 (a) 23
(b) Substitutable; a rise in P A leads to an increase in Q.
(c) 6
5 a = −6, b = 720
6 (a) 20, 10, 45; line passes through these three points.
(b) Line passing through (50, 0) and (0, 50)
Q = 20, P = 30
(c) Price increases; quantity increases.
8 P 1 = 40, P 2 = 10; Q 1 = 30, Q 2 = 55
9 (a) Q = 30
(b) Substitutable; e.g. since coefficient of Pr is positive.
(c) P = 14
(d) (i) slope = − 20 , intercept = 135
(ii) slope = 1
, intercept = 6.
Exercise 1.5*
1 (a) As P S rises, consumers are likely to switch to
the good under consideration, so demand
for this good also rises: that is, the graph shifts
to the right.
(b) As P C rises, demand for the bundle of goods
as a whole is likely to fall, so the graph shifts
to the left.
(c) Assuming that advertising promotes the good
and is successful, demand rises and the graph
shifts to the right. For some goods, such as drugs,
advertising campaigns are intended to discourage
consumption, so the graph shifts to the left.
(^2) m 5 2 3 c 5 9 2
3 0 and 30
4 (1) P = 30, Q = 10
(2) New supply equation is 0.85 P = 2 Q S + 10;
P = 33.6, Q = 9.28.
5 (a) 53, 9 (b) $
6 P 1 = 20, P 2 = 5, P 3 = 8; Q 1 = 13, Q 2 = 16, Q 3 = 11
7 Change supply equation to P = 2 Q S + 40 + t.
In equilibrium
− 3 Q + 60 = 2 Q + 40 + t − 5 Q = − 20 + t
Q 5 4 2 t 5
Substitute to get 3 t
P 5 48 1
(a) t = 5; firm pays $2 (b) P = 45, Q = 5
8 (a) a > 0, b > 0, c < 0, d > 0
(b) (^) Q 5 d^^2 b a 2 c
ad 2 bc a 2 c
and P 5
Section 1.
Practice Problems
1 (a) Q = 8
(b) Q = 2 P − 26 (multiply out brackets)
(c) Q = 2 × 17 − 26 = 8
2 (a) x 5 y 6
(b) x (^5 17) y^1 (^1 15 1 71) y^ y
3 (a) (^) x 5 1 1 a y 1 2 c
(b) (^) x 5 22 2 4 y y 2 1
Exercise 1.
1^1 P 2 4; 22
Q 5
2 (a) y = 2 x + 5 (b) y = 2( x + 5)
(c) 5 x^2
y 5 (d) y = 2( x + 4)^2 − 3
3 (a) multiply by 5 add 3
(b) add 3 multiply by 5
(c) multiply by 6 subtract 9
(d) square multiply by 4 subtract 6
(e) divide by 2 add 7
(f) reciprocate multiply by 2
(g) add 3 reciprocate
4 (a) 1 x 5 9 ( y 1 6) (b) x = 3 y – 4
(c) x = 2 y (d) x = 5( y − 8)
(e) y
x 5 12 2 (f) y
x 5 41 7
5 (a) Q a
b a
P 5 2 (b) b^^1 I 1 2 a
Y 5
(c) aQ
b a
P 5 1 2
y 1 2
x 5 3
7 (a) D 5 HQ
2 2 R
(b) H 5 Q^2
2 DR
Exercise 1.6*
1 (1) (a) multiply by 9 add 1
(b) multiply by − 1 add 3
(c) square multiply by 5 subtract 8
(d) multiply by 3 add 5 square root
(e) square add 8 reciprocate multiply by 4
(2) (a) y^^2 x (^5 ) (b) x = 3 − y
(c) y^^1 x 5 6 (^5) (d) y
x (^53)
(e) y
x 5 6 2 8
2 (a) c^^2 a x (^5) b (b) a
(^2 1) b a 1 1
x 5
(c) x = ( g − e )^2 − f (d) ma
2 x (^5) b 2 1 n
(e) n
2 x (^5) m 21 m (f) a
(^2 1) b 2 b 2 a
2 x 5
3 V^^1
t (^5) V 2 5 ; 11
4 S
P
r 5 100 n 2 1
5 (a) G = Y (1 − a + at ) + aT − b − I
(b) a
G 1 b 1 I 2 Y (1 2 a 1 at ) T 5
(c) aY
G 1 b 1 I 2 Y 1 aY 2 aT t 5
(d) T 2 Y 1 tY
a 5 G^^1 b^^1 I^^2 Y
50
50
100
100
150
150
200 45 degree line C 1 I line
200 Y
Figure S1.
Y = 160
If MPC decreases, the slope decreases as
indicated by the dashed line. The point of
intersection shifts down the 45° line, showing
that the equilibrium value of Y decreases.
7 (a) The coefficient of P 2 in the demand equation for
good 1 is positive, indicating that demand for
good 1 rises as the price of good 2 goes up.
Similarly, for the second equation. Hence the
goods are substitutable.
(b) P 1 = 15, P 2 = 8; Q 1 = 98, Q 2 = 157
(c) The price of good 1 reduces by $3.15 and the
price of good 2 reduces by $0.35.
8 (a) 3 f + 4 s ≤ 3000 ; 640 kg
(b) (i) 3
(ii) 5
(iii) Q 5 36 2 5 P
2 P
(c) S 5
Y 1 2
2 Y 2 100
9 (a) x 5 y 5
2 9 2 m
3 2 2 n
2 ( 9 2 m )
6 n 2 m , ; m = 9
(i) Infinitely many solutions
(ii) No solutions
(b) P 1 = 12, P 2 = 7, P 3 = 8 10 Y = 3160, r = 2.
Figure S1.22 (not to scale) shows the IS and LM curves
intersecting at the equilibrium point (2.4, 3160). The
dashed line shows the new IS curve after a reduction
in MPC. The point of intersection has moved down
The graphs are sketched in Figure S1.20.
5
50
10
100
15
150 200 250 300 350
20
25
P
Q
Figure S1.
(b) P = 18, Q = 130 (c) P < 18
(d) The demand curve is unchanged and the supply
curve shifts upwards by 5 units.
P = 20, Q = 100
3 (a) (1) is demand and (2) is supply.
(b) slope = −1/3, vertical intercept = 16 (c) P = 3.6, Q = 37.
4 (a) 1850
(b) 5
5 (a) 12
(b) 20
(c) K 5 2 QL
L 2 Q
(b) 12 miles
(c) EatMeNow: y = 0.5 x + 9 , Deliver4U: y = 1.25 x
2 (a) P 5 10 15 20 25 Q D 325 250 175 100 25 Q S 0 50 100 150 200
6 Y 0 100 200
C + I^40 115
The lines are sketched in Figure S1.21.
(b) x 0 1 2 3 4 5 6 f ( x ) (^) − 9 − 4 − 1 0 − 1 − 4 − 9
The graph is sketched in Figure S2.2.
2800
3400
(2.4,3160)
LM
IS
34 r
Y
Figure S1.
ChApter 2
Section 2.
Practice Problems
1 (a) x = ± 10 (b) x = ± 2
(c) x = ±1.73 (d) x = ±2.
(e) No solution (f) No solution
(g) x = 0
2 (a) x = 10 and 1
2 (b) 3 2
x 5 2
(c) No solution (d) x = 2 and 3
3 (a) x^ − 1 0 1 2 3 4
f ( x ) (^21 5) − 3 − 3 5 21
The graph is sketched in Figure S2.1.
1 2
20 15 10 5
(^3 4) x
f ( x ) 5 4 x^2 2 12 x 1 5
y
21 25
Figure S2.
and to the left indicating a decrease in the equilibrium
values of both Y and r.
1 2 3 4 5 6 x
f ( x ) 5 2 x^2 1 6 x 2 9
y
22 24 26 28 210 212
Figure S2.
(c) x (^) − 2 − 1 0 1 2 3 4 f ( x ) (^) − 22 − 12 − 6 − 4 − 6 − 12 − 22
The graph is sketched in Figure S2.3.
1 2 3 4
f ( x ) 5 2 2 x^2 1 4 x 2 6
y
(^2221) x 25 210 215 220 225 230
Figure S2.
4 (a) The graph is sketched in Figure S2.4.
6
f ( x ) 5 2 x^2 2 11 x 2 6
y
x
26
1
( 114 ,^2 1698 )
(^22)
Figure S2.
8 (a) x < 0, x > 3 (b) x ≤ −1, x ≥ 1
(c) − 4 < x < 2
9 Q = 4, P = 36
10 P = 22, Q = 3
11 (a) $277 (b) 85
Exercise 2.1*
1 (a) ± 13 (b) −3, 13 (c) −2, 9
2 − 7 d , d
3 (a) 3, − 8 (b) 2
(c) 0, 3
(d) 1 6
(twice)
(e) 2, −1, 4
4 (a) 7, 8 (b) 0.22, 2.
(c) ± 3 (d) 7 (twice)
(e) No solutions (f) 10, 19
5 (a) x ≤ −8, x ≥ 8 (b) 1 ≤ x ≤ 9
(c) 1 2
27 , x , 2 (d) 5 3
21 # x # (e) x = − 1
6 c = 12; 6
7 k = 27
8 (a) x ≤ −3, x ≥ 4 (b) − 1 < x < 2
(c) x ≤ 1, 2 ≤ x ≤ 3 (d) 2 ≤ x < 3, x > 5
11 (a) P = 18 (b) B = 15
12 (b) between $18 and $40 (c) $
13 P = 5, Q = 65
Section 2.
Practice Problems
1 TR = 1000 Q − Q^2
The graph is sketched in Figure S2.8.
Q = 500 and P = 500
Q
TR 5 1000 Q 2 Q^2
0 1000
TR (500, 250 000)
Figure S2.
2 TC = 100 + 2 Q
Q
AC 5 1 2
The graph of the total cost function is sketched in
Figure S2.9.
Q
TC
slope 5 2
intercept 5 100
Figure S2.
The graph of the average cost function is sketched in
Figure S2.10.
AC (^5 100) Q 1 2
AC 12 10 8 6 4 2
(^50 100 150 200) Q
Figure S2.
3 π = − Q^2 + 18 Q − 25
The graph of the profit function is sketched in
Figure S2.11.
31
4 14 1.52 16.
(9, 56) (^) p 5 2 Q (^2 1 18) Q 2 25
Q
p
225
Figure S2.
(a) Q = 4 and 14 (b) Q = 9; π = 56
Exercise 2.
1 (a) P = 50; TR = 500
(b) TC = 150
(c) π = 350
2 (a) 4 Q
(b) 7
(c) 10 Q − 4 Q^2
The graphs are sketched in Figures S2.12, S2.13 and
S2.14.
TR TR 5 4 Q
Q
Figure S2.
TR TR 5 7 7
Q
Figure S2.
TR (1.25, 6.25)
0 2.5 Q
TR 5 10 Q 2 4 Q^2
Figure S2.
3 (a) P = 50 − 4 Q
(b) Q
P 510
4 TC = 500 + 10 Q ; 500
AC 5 Q 1 10
The graphs are sketched in Figures S2.15 and S2.16.
TC
Q
slope 5 10
intercept 5 500
Figure S2.
Q
AC
10
AC (^5 500) Q 1 10
Figure S2.
5 TC = Q^2 + Q + 1;
Q
AC 5 Q 1 1 11
The graphs are sketched in Figures S2.17 and S2.18.
TC
1
Q
TC 5 Q^2 1 Q 1 1
Figure S2.
(b) f ( K , L ) = 50 K 1/4 L 3/ f (l K , l L ) = 50(l K )1/4(l L )3/
= 50 l1/4 K 1/4l3/4 L 3/4^ (rule 4)
= (l1/4l3/4)(50 K 1/4 L 3/4)
= l^1 f ( K , L ) (rule 1)
Constant returns to scale.
5 (1) (a) 3 (b) 2 (c) 1 (d) 0
(e) − 1 (f) − 2
(2) Same as part (1).
6 (a) log b
xz y (b) log b ( x^4 y^2 )
7 (a) x = 1.77 (b) x = 1
Exercise 2.
1 (a) 64 (b) 2 (c) 1/3 (d) 1
(e) 1 (f) 6 (g) 4 (h) 1/
2 (a) a^11 (b) b^5 (c) c^6 (d) x^2 y^2
(e) x^3 y^6 (f) y −^4 (g) x^4 (h) f^7 (i) y^3 ( j) x^5
3 (a) x
(^12) (b) x −^2 (c) x
(^13) (d) x −^1 (e) x 212 (f) x
(^32)
4 (a) 3600 (b) 200 000
5 The functions in parts (a) and (b) are homogeneous
of degree 7/12 and 2, respectively, so (a) displays
decreasing returns to scale and (b) displays increasing
returns to scale. The function in part (c) is not
homogeneous.
6 (a) 2 (b) − 1 (c) − 3 (d) 6 (e) 1 / 2 (f) 0
7 (a) 2 (b) 1 (c) 0 (d) 1 /2 (e) − 1
8 (a) log b ( xz ) (b) log b x
3 y^2
(c) log b y z^3
9 (a) 2 log b x + log b y (b) log b x − 2 log b y
(c) 2 log b x + 7 log b y
10 (a) 1.29 (b) 1.70 (c) 6.03 (d) 8.
11 (1) (a) 5 (b) 1 2
(2) log b x
2 y^4 (3) 69.
12 (1) (a) 4 (b) − 2 (c) 2
(2) (a) x^3 y (b) x^15 y^5 (c) x^2 y^2
13 (a) 98, 115, 125, 134, 140, 146
(b) The graph is sketched in Figure S2.20.
Figure S2.
The number of complaints increases but at a
decreasing rate.
Exercise 2.3*
1 (a) 8 (b) 1/32 (c) 625 (d) 2 1 4
(e) 2/
2 (a) y^2 (b) xy^2 (c) x^4 y^2 (d) 1 (e) 2 (f) 5 pq^2 3 (a) x −^7 (b) x 1/4^ (c) x −3/2^ (d) 2 x 11/2^ (e) 8 x −4/ 4 3 x^3 y^7 5 A [ b (l K )a^ + (1 − b )(l L )a]1/a
= A [ b la K a^ + (1 − b )la L a]1/a^ (rule 4)
= A [(la)( bK a^ + (1 − b ) L a)]1/a^ (factorise)
= A (la)1/a^ [ bK a^ + (1 − b ) L a]1/a^ (rule 4)
= l A [bKa^ + (1 − b ) L a]1/a^ (rule 3)
so f (l K , l L ) = l^1 f ( K , L ) as required.
6 (a) 2/3 (b) 3 (c) 1 /
7 (a) log b (1) (b) log b
x^3 y^2 (c)^ log b
x^5 y z^2 (d) log b ( b^2 x^3 ) 8 (a) 2 log b x + 3 log b y + 4 log b z (b) 4 log b x − 2 log b y − 5 log b z
(c) 1 2
log b x 2 log b y 2 log b z
N
140
120
100
80
60
40
20
1 2 3 4 5 6 t
9 (a) − q (b) 2 p + q (c) q − 4 r (d) p + q + 2 r
10 (a) 78.31 (b) 1.48 (c) 3 (d) 0.
11 (a) x ≤ 0.386 (3 dp)
(b) x > 14.
12 x = 3
13 (2) 2 3
; constant returns to scale.
14 (1) (a)
3 (b)^
(2) 1 x^2 7
y 5
15 log^10 x^2 2 log 10
5 log (^10)
5 log (^10)
y 2 log 10 10
2 1
x^2 y
x^4 y
16 (a) L 5
Q
AK a
1/b (b) L 5 ( Q A^ /^ )
a (^2) bK a 1 2 b
1/a
17 (a) 1 (b) 2 (c) n (d) 3
Section 2.
Practice Problems
1 x (^) − 3 − 2 − 1 0 1 2 3
3 x^ 0.04 0.11 0.33 1 3 9 27 3 − x^^27 9 3 1 0.33^ 0.11^ 0.
The graphs of 3 x^ and 3− x^ are sketched in Figures S2.
and S2.22, respectively.
y
1 x
3
6
9
12
15
18
21
24
27
23 22 21 2 3
y 5 3 x
Figure S2.
x
y
21 1
3
6
9
12
15
18
21
24
27
23 22 2 3
y 5 32 x
Figure S2.
2 (a) 2. 718 145 927, 2.718 268 237, 2.718 280 469
(b) 2.718 281 828 ; values in part (a) are getting closer
to that of part (b).
3 (1) (a) 0.07% (b) 1.35% (c) 18.44% (d) 50.06% (2) 55%
(3) A graph of y against t , based on the information
obtained in parts (1) and (2), is sketched in
Figure S2.23. This shows that, after a slow start,
microwave ownership grows rapidly between
t = 10 and 30. However, the rate of growth then
decreases as the market approaches its saturation
level of 55%.
t
y 55 50 45 40 35 30 25 20 15 10 5 5 10 15 20 25 30 35 40 45 50
55 y (^51 1) 800e 2 0.3 t
Figure S2.
4 (a) 2 ln a + 3 ln b (b) ln^
x 1/ y^3
5 (a) $5 million and $3.7 million
(b) 4 years
6 (a) 3 y^2 + 13 y − 10
(b) Put y = e x^ in part (a) to deduce −0.405.
7 (a) b
x a
(^1) ln y 5 (b)^
x 5 ln(e y^ 2 3)
8 (a) 6 (b) −1 or 2 (c) 25 (d) 0.26 (e) ±1.
9 P^^5 k 1 1 k 2
ln( A / B )
Multiple Choice Questions
1 B 2 D 3 C
4 B 5 C 6 D
7 C 8 A 9 A
10 B 11 D 12 E
13 D 14 B 15 A
Examination Questions
1 (a) 2000
(b) 100
(c) 64
(d) Show f (l K , l L ) = l7/6^ f ( K , L ); increasing returns
to scale
2 π = − Q^2 + 6 Q − 8. A graph of the profit function is
sketched in Figure S2.
5 (a) (i) x
6
y
log a
(ii) − 2
(b) Each year the price is multiplied by the same
factor, 1.1; A = 1000; b = 1.
(i) $
(ii) At the beginning of Year 22
6 (a) TC = 14 + 2 Q ; AC =
Q
1 2. The graphs are
sketched in Figures S2.27 and S2.28.
y 1
2
210
29
28
27
26
25
24
23
22
21 1 2 3 4 5 6 x
Figure S2.
(a) Q = 2, 4
(b) 1
(c) Shifts vertically upwards
3 (a) P = 100, Q = 6
(b) P = 101.79, Q = 5.88; consumer pays $1.79 and
producer pays $3.
4 (a) TR = 10 Q − Q^2
(b) 5 ≤ P ≤ 7
TC
14
y
Q
Figure S2.
AC
2
Q
Figure S2.
(b) When Q = 0 , there is no revenue and the only
costs are the fixed ones which are given to be 14,
so the profit is 0 − 14 = − 14. Substituting Q = 0
into π = aQ^2 + bQ + c gives π = c , so c must
equal − 14.
(c) a + b = 8, 6 a + b = 3; a = −1, b = 9
(d) Q = 2, 7; max profit is 6.
(e) TR = − Q^2 + 11 Q ; P + Q = 11
7 (a) (i) a
5 b 3
(ii) 4 x^2 y^4 (iii) 6
(iv) log b ( xy^2 z^2 )
(b) −1.
(c) (i) 659,
(ii) 3.
(iii) The sales graph is sketched in Figure S2.29.
7 (a) 5.7% increase (b) 18.4% increase
(c) 12.0% increase
8 The complete set of ‘constant Year 2 prices’ is listed
in Table S3.1.
During Year 1/2 salaries remain unchanged in real
terms. However, since Year 2 salaries have outpaced
inflation with steady increases in real terms.
Table S3.
Year 1 2 3 4 5 Real salaries 18.1 18.1 19.0 21.7 23.
Exercise 3.
1 (a) 7 20
(b) 22 25
(c) 2 1 2
(d) 7 40
(e) 1 500 2 (a) 1.2 (b) 7.04 (c) 2190.24 (d) 62. 3 (a) 60% (b) 22 4 (a) 1.19 (b) 3.5 (c) 0.98 (d) 0.
5 (a) 4% increase (b) 42% increase
(c) 14% decrease (d) 245% increase
(e) 0.25% increase (f) 96% decrease
6 (a) $18.20 (b) 119 244 (c) $101. (d) $1610 (e) $15 640 7 35% 8 (a) $15.50 (b) $10.54 (c) 32% 9 $862.
10 (a) $26 100 (b) 31% (nearest percentage)
11 (a) 37.5% increase (b) 8.9% increase
(c) 6.25% decrease
12 $11.6 million
Prices rise consistently over the past five years at an
steadily increasing rate.
14 (a) Jan (b) 4800 (c) 133
(2) (a) 30% (b) 52.3% (c) 13.1% (d) 9.4% (3) (a) 25% (b) 44% (c) 10.6% (d) 11.1%
(4) Public transport costs have risen at a faster rate
than private transport throughout the period
1995–2010. However, for the final five years there
are signs that the trend has stopped and has
possibly reversed.
16 964, 100, 179, 750; e.g. seasonal variations.
17 (a) 83.3, 100, 91.7, 116.7, 125, 147. (b) 64.8 (c) Year 5
2 million
S
t
Figure 2.
8 (a) A = 50,000; b = 0.95; A is the initial price
(b) r = −0.
(c) Use rules to logs to obtain ln V = −0.0513 t + ln 50000
so it is a line with gradient −0.0513 and vertical
intercept 10.8.
9 (a)
Q
1 0.5 1 0.04 Q ;
Q
(b) A = IT; B = Printing; C = Retail
(c) 2.5 < Q < 10
ChApter 3
Section 3.
Practice Problems
1 (a) $0.29 (b) $937.50 (c) $139.
2 (a) 10% (b) $1564 (c) $
3 (a) $7.345 million (b) $76 billion (c) 7%
4 (a) 8750 (b) 750 (c) 80%
5 (a) 82% increase (b) 58% decrease
(c) 45% decrease
