Basic Mathematics for Economists - Rosser - Chapter 5 ppsx
Basic Mathematics for Economists - Rosser - Chapter 5 ppsx
... price
p
∗
s
= p
d
(55 + 4q)(1 + t) = 3 75 − 2.5q
55 + 55 t + 4q + 4qt = 3 75 −2.5q
6.5q + 4qt = 320 − 55 t
q(6 .5 + 4t) = 320 − 55 t
q =
320 − 55 t
6 .5 + 4t
This reduced form equation for equilibrium q ... + 85. 5 = 120
x = 120 −109 .5
x = 10 .5
Therefore, the solutions are x = 10 .5, y = 2,z = 28 .5.
Example 5. 11
Solve for x, y and z in the following set of simultaneous e...
Basic Mathematics for Economists - Rosser - Chapter 12 ppsx
... rule
d
2
TR
dq
2
= u
dv
dq
+ v
du
dq
= ( 650 − 0.25q)
0 .5
(−0.6 25) + ( 650 − 0.625q)(−0.1 25) ( 650 − 0.25q)
−0 .5
=
( 650 − 0.25q)(−0.6 25) + ( 650 − 0.625q)(−0.1 25)
( 650 − 0.25q)
0 .5
(2)
Substituting the value q = ... =−0.3 75( 650 −0.25q)
0 .5
and also
dv
dq
= 1
Therefore, using the product rule
dTR
dq
= u
dv
dq
+ v
du
dq
= ( 650 − 0.25q)
1 .5
+ q(−0.3 75) ( 650 − 0.25q)
0 .5
=...
Basic Mathematics for Economists - Rosser - Chapter 1 pptx
... 7? 51 ? 39? It all depends on which order you perform the
calculations and the type of calculator you use.
There are set rules for the order in which basic arithmetic operations should be performed,
whichareexplainedinChapter2.Nowadays,theseareprogrammedintomostcalculators
but ... quantification
of economic predictions requires the use of mathematics.
Although non-mathematical economic...
Basic Mathematics for Economists - Rosser - Chapter 2 potx
... elasticity
e =
150
600 + 750
150
350 + 200
=
150
1, 350
150
55 0
=
150
1, 350
×
55 0
150
=
1
27
×
11
1
=
11
27
£
D
0 40 100 12060 80
18
20
6
3
Quantity
Price
9
15
12
Figure 2.2
© 1993, 2003 Mike Rosser
Test ... in the form with a superscript, e.g. 6
0 .5
,
so that the rules for multiplying powers can be applied.
Example 2.43
47
0 .5
× 47
0 .5
= 47
Example 2.44
15 ×...
Basic Mathematics for Economists - Rosser - Chapter 3 pps
... 123 456 789101112
Height (cm) 178 1 75 170 166 168 1 85 169 189 1 75 181 177 180
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 ... Exercise 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 = 1,0...
Basic Mathematics for Economists - Rosser - Chapter 4 potx
... Rosser
TestYourself,Exercise4.10
Fortheproductionfunctionsbelow,assumefractionsofaunitofKandLcanbe
used,and
(a)deriveafunctionfortheisoquantrepresentingthespecifiedoutputlevelinthe
formK=f(L)
(b)findthelevelofKrequiredtoachievethegivenoutputlevelifL=100,and
(c)saywhattypeofreturnstoscalearepresent.
1.Q=9K
0 .5
L
0 .5
,Q=36
2.Q=0.3K
0.4
L
0.6
,Q=24
3.Q=25K
0.6
L
0.6
,Q=800
4.Q=42K
0.6
L
0. 7...
Basic Mathematics for Economists - Rosser - Chapter 6 pps
... 24 .5 -6 82 35. 5 -2 97
10 3 -3 7 14 -5 98 25 -6 75 36 -2 68
11 3 .5 -7 3 14 .5 -6 12 25. 5 -6 67 36 .5 -2 38
12 4 -1 08 15 -6 25 26 -6 58 37 -2 07
13 4 .5 -1 42 15. 5 -6 37 26 .5 -6 48 37 .5 -1 75
14 5 -1 75 16 -6 48 ... 11 -4 93 22 -7 02 33 -4 27
5 0 .5 158 11 .5 -5 13 22 .5 -7 00 33 .5 -4...
Basic Mathematics for Economists - Rosser - Chapter 7 pps
... 0.613913 0 .55 83 95 0 .50 8349 0.463193 0.422411 0.6 755 64 0.613913 0 .55 83 95 0 .50 8349 0.463193 0.422411
11 0.64 958 1 0 .58 4679 0 .52 6788 0.4 750 93 0.428883 0.38 753 3 0.64 958 1 0 .58 4679 0 .52 6788 0.4 750 93 0.428883 ... 0.667263 0 .54 5061 0.363699 0.017497 0.006363
1. 75 0.812 057 0. 659 438 0 .53 550 1 0. 353 130 0.0 155 50 0.0 054 91
1.80 0.807284 0. 651 708 0 .52 6114 0....
Basic Mathematics for Economists - Rosser - Chapter 8 doc
... −0.5q
0.3q
2
− 0.5q − 48 = 0
Using the formula for the solution of quadratic equations
q =
−(−0 .5) ±
(−0 .5)
2
− 4 ×0.3 × (−48)
2 × 0.3
=
0 .5 ±
√
0. 25 + 57 .6
0.6
=
0 .5 ±
√
57 . 85
0.6
Disregarding the negative ... demand function
q = 60 − 2p
0 .5
2p
0 .5
= 60 − q
p
0 .5
= 30 − 0.5q
p = (30 − 0.5q)
2
= 900 − 30q + 0.25q
2
© 1993, 2003 Mike Rosser
Solution
Slope =
dy
dx
= 3...
Basic Mathematics for Economists - Rosser - Chapter 9 pptx
... Rosser
Thereforethesecond-orderconditionforaminimumvalueofACissatisfiedwhenqis5.
TheactualvalueofACatitsminimumpointisfoundbysubstitutingthisvalueforqinto
theoriginalACfunction.Thus
AC=25q
−1
+0.1q
2
=
25
5
+0.1× 25= 5+2 .5= 7 .5
TestYourself,Exercise9.3
Findwhetheranystationarypointsexistforthefollowingfunctionsforpositive
valuesofq,andsaywhetherornotthestationarypointsareattheminimumval...