Basic Mathematics for Economists - Rosser - Chapter 6 pps
Basic Mathematics for Economists - Rosser - Chapter 6 pps
... 26. 5 -6 48 37.5 -1 75
14 5 -1 75 16 -6 48 27 -6 37 38 -1 42
15 5.5 -2 07 16. 5 -6 58 27.5 -6 25 38.5 -1 08
16 6 -2 38 17 -6 67 28 -6 12 39 -7 3
17 6. 5 -2 68 17.5 -6 75 28.5 -5 98 39.5 -3 7
18 7 -2 97 18 -6 82 ... -5 67 24 -6 88 35 -3 25
9 2.5 0 13.5 -5 83 24.5 -6 82 35.5 -2 97
10 3 -3 7 14 -5 98 25 -6 7...
Basic Mathematics for Economists - Rosser - Chapter 3 pps
... 170 166 168 185 169 189 175 181 177 180
Weight (kg) 72 68 58 52 55 82 55 86 70 71 65 68
© 1993, 2003 Mike Rosser
Example 3.18
Simplify (6 − 5x)(10 − 2x + 3y).
Solution
Multiplying out gives
60 ... 3.7
1. Solve for x when 16x = 2x + 56.
2. Solve for x when
14 =
6 + 4x
5x
3. Solve for x when 45 = 24 +3x.
4. Solve for x if 5x
2
+ 20 = 1,000.
5. If q = 560 − 3p solve for p w...
Basic Mathematics for Economists - Rosser - Chapter 5 ppsx
... check that the price of £12. 16 for the first 60 units corresponds to point A on the
demand schedule in Figure 5 .6 since
p = 16 − 0. 064 q = 16 −0. 064 (60 ) = 16 −3.84 = £12. 16
If the firm wishes to sell ... 120)
q = 6MC − 260
q + 260
6
= MC (1)
Since
p = 580 −0.3q
MR = 580 − 0.6q (2)
To maximize profits MC = MR and so equating (1) and (2)
q + 260
6
= 580 − 0.6q
q + 260 = 3,480...
Basic Mathematics for Economists - Rosser - Chapter 7 pps
... 0.7 462 15 0.704 961 0 .66 6342 0 .63 017 0.5 962 67
7 0.759918 0.71 068 1 0 .66 5057 0 .62 275 0.58349 0.547034 0.759918 0.71 068 1 0 .66 5057 0 .62 275 0.58349 0.547034
8 0.73 069 0 .67 6839 0 .62 7412 0.582009 0.540 269 ... 0.009908
1 .60 0.8 265 62 0 .68 3204 0. 564 711 0.385813 0.0221 56 0.008584
1 .65 0.82 169 6 0 .67 5185 0.554797 0.374590 0.01 968 9 0.007375
1.70 0.8 168 61 0 .66 7 2...
Basic Mathematics for Economists - Rosser - Chapter 11 pps
... 40 − 22.5(275, 567 .6) K
−2.5
= 0 (3)
40 =
22.5(275, 567 .6)
K
2.5
K
2.5
=
22.5(275, 567 .6)
40
= 155,0 06. 78
K = 119. 162 68
Substituting this value into (1) gives
L =
275, 567 .6
(119. 162 8)
1.5
= 211.84478
This ... 4K
0 .6
(1. 562 5K)
0.5
= 0
200 = 4K
0 .6
(1. 562 5)
0.5
K
0.5
200
4(1. 562 5)
0.5
= K
1.1
40 = K
1.1
K =
1.1
√
40 = 28 .60 3434
Substituting this value into (4)
1. 562 5...
Basic Mathematics for Economists - Rosser - Chapter 12 ppsx
... (2)
d
2
Q
dK
2
=
( 160 − 8K)
0 .6
K
0 .6
(− 76. 8) − ( 768 − 76. 8K)
96 − 9.6K
K
0.4
( 160 − 8K)
0.4
( 160 − 8K)
1.2
K
1.2
=
( 160 − 8K)K(− 76. 8) − 76. 8(10 − K)9 .6( 10 − K)
( 160 − 8K)
1 .6
K
1 .6
At the stationary ... 4.8K
−0 .6
and
dv
dK
= 0.4( 160 − 8K)
−0 .6
(−8)
=−3.2( 160 −8K)
−0 .6
Therefore,
dQ
dK
= 12K
0.4
(−3.2)( 160 −8K)
−0 .6
+ ( 160 − 8K)
0.4
4.8K
−0 .6
=
−38.4K +(...
Basic Mathematics for Economists - Rosser - Chapter 14 pps
... 1.75 years, r = 2.5%= 0.025, A = 56 (e billion). Therefore, the
final value of GNP will be
y = Ae
rt
= 56e
0.025(1.75)
= 56e
0.04375
= 58.504384
Thus the forecast for GNP is e58,504,384,000.
So far ... example, when t = 3 then
y
3
= 22.5e
6( 3)
− 4.5 = 22.5e
18
− 4.5 = 22.5 (65 ,65 9, 969 ) − 4.5 = 1,477,349,303
Before we investigate the usefulness of this method for the analysis of dyna...
Basic Mathematics for Economists - Rosser - Chapter 1 pptx
... which basic arithmetic operations should be performed,
whichareexplainedinChapter2.Nowadays,theseareprogrammedintomostcalculators
but not some older basic calculators. If you only have an old basic ... economics degree courses gave up mathematics at school
at the age of 16, many of them because they thought that they were not good enough at
mathematics to take it for A-level. Howe...
Basic Mathematics for Economists - Rosser - Chapter 2 potx
... Example 2.32
6
6
6
3
=
6 × 6 × 6 × 6 × 6 6
6 × 6 × 6
= 6 × 6 × 6 = 6
3
= 2 16
In the two examples above the multiplication and division processes ... 123
6
=
3. 6
4
÷ (6
2
× 6) =
4. (−2)
3
× (−2)
3
=
5. 1.42
4
× 1.42
3
=
6. 9
5
× 9
−3
× 9
4
=
7. 8 .67 3
3
÷ 8 .67 3
6
=
8. ( 6)
5
× ( 6)
−3
=
9. (−8.52)
4
× (−8.52)
−1
=
10. (−2.5)
−8
+ (0.2)
6
× ... of...
Basic Mathematics for Economists - Rosser - Chapter 4 potx
... 3
1
3
MR
1
(1)Q
2
= 60 − 6
2
3
MR
2
(2)
© 1993, 2003 Mike Rosser
0
K
Q
1
Q
2
Q
3
L
Figure 4.21
Table 4.7
KLK
0.5
L
0.5
Q
64 4 8 2 320
16 16 4 4 320
464 2 8320
2 56 1 16 1 320
1 2 56 1 16 320
Assume that ... decrease of £400, since 60 0 = 1.5 × £400. Therefore
the fall in C is 1.5 ×£240 = £ 360 .
C (£)
0(£)
C = 200 + 0.6Y
Y
900
1,00 060 0
66 0
300
Figure 4.7
© 1993, 2003 Mike R...