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- QUESTION
Mathematical Economics---Final Exam
Mathematical Economics--- Final Exam
Name :
- Find the differential dy,given(10 marks, 5marks each):
(a)y=-x(2x2+3)
(b)y=(X-8)(7x+8)
2.Given S1={1,2,3},S2={a,b},and S3={m,n},find the Cartesian products: (21 marks, 7 marks each)
- S1XS2 (b)S2XS3 (c)S3XS1
3.Find the value of the following factorial expressions: (25 marks, 5marks each)
- 5!
- 6!
- 4!/3!
- 6!/4!
- (n+2)!/n!
4.Find the stationary values of the following functions: (10marks, 5arks each)
- Y=x4
- Y=-x3
Determine by the Nth-derivative test whether they represent relative maxima, relative minima, or inflection points
5.Let the demand and supply functions be as follows: (20 marks, 10marks each)
- Qd=51-3P (b) Qd=30-2P
Qs=6P-13 Qs=-5+5P
Find P*and Q* by elimination of variables.(Use fractions rather than decimals.)
6.Find the characteristic vectors of the matrix (4 2) (14 marks)
( 2 1)
| Subject | Economics | Pages | 8 | Style | APA |
|---|
Answer
Mathematical Economics--- Final Exam
- Find the differential dy, given
(a) y=-x (2x2+3)
(b) Y=(X-8) (7x+8)
- a) Y = -2x3 + 3x
dy/dx = -6x2 + 3
- b) y = (x-8) (7x+8)
y= 7X2 – 48X -64
dy/dx = 14x – 48
- Given S1={1,2,3},S2={a,b},and S3={m,n},find the Cartesian products
Given S1 = {1, 2, 3}, S2 = {a, b}, S3 = {m, n}
In this case, S1, S2 S3 = Ø
- S1 X S2
From principles, AXB = {(a, b): aϵA, bϵb}
For non-empty sets A and B
Thus for this case, for non-empty sets S1 and S2
The cartesian product {1,2,3} X {a,b} = {(1,a), (1,b),(2,a), (2,b), (3,a)(3,b)}
- S2X S3,
Following as in a) above, both are non-empty sets, hence
{a, b} X {m, n} = {(a, m), (a, n), (b, m), (b, n)}
- S3X S1, {m,n}X {1,2,3} = {(m,1), (m,2),(m,3), (n,1),(n,2), (n,3)}
- Find the value of the following factorial expressions: (25 marks, 5marks each)
- 5!
- 6!
- 4! /3!
- 6! /4!
- (n+2)! /n!
- a) 5!
n! = n (n-1)!
Hence
5! = 5X4X3X2X1
= 120
- b) 6!
From n! =n (n-1)!
6! = 6X5!
6X 5!
6X120
720
- 4! /3!
From n! = 4X3! /3!
=4
- 6! /4!
From principle, n! = n (n-1)!
6 X5 X4! /4!
=30
- (n+2)! /n!
From principles, n! = n (n-1)!
(n +2) (n+1) n! /n!
=n2 + 3n + 2
- Find the stationary values of the following functions
- Y=x4
- Y=-x3
Determine by the Nth-derivative test whether they represent relative maxima, relative minima, or inflection points
Stationary values
Principles, if dy/dx = 0 => stationary point, first order derivative = 0
To determine maxima, minima, or point of local inflection, we use the second order derivative by applying the following principles d2y/dx2
d2y/dx2 <0 at that point => local maxima,
d2y/dx2 >0 at that point => local minima,
d2y/dx2 =0 at that point => local point of inflection,
Hence for given equations below
- y = x4
dy/dx = 0 at a stationary point =>
dy/dx = 4x3 = 0
At x=0, y=0 hence at point (0, 0)
d2y/dx2 = 12x2
At the point x = 0, d2y/dx2 =0
Implies point of inflection
- –x3
dy/dx = -3x2
Stationary point, -3x2 =0, => x=0
d2y/dx2 = -6x
At x=0, d2y/dx2 = 0
From principles, d2y/dx2 = 0 at this point hence point of inflection
- Let the demand and supply functions be as follows
- Qd=51-3P (b) Qd=30-2P
Qs=6P-13 Qs=-5+5P
Find P*and Q* by elimination of variables. (Use fractions rather than decimals.)
- a) Qd = 51 – 3p ---i
Qs = 6p - 13 -----ii
By elimination, multiply i X 2 and ii X1
Thus
2Q = 102 – 6p ----iii
Q = -13 +6p ------IV
Adding the two
3Q = 89
Q = 89/3
= 29 2/3,
Substituting,
89/3 = 51 – 3p
89 = 153 – 9p
9p = 64
P = 64/9
P= 7 1/9
- Qd =30 – 2p ---i
Qs = -5 +5p ---ii
iX 5
iiX2
5Q = 150 -10p --iii
2Q = -10 +10p—iv
Add iii + iv
7Q = 140
Q= 20
20 = 30 – 2P
2P = 10
P = 5
- Find the characteristic vectors of the matrix )
Def. )=
() ()-(2) (2)
4-4λ-λ+λ-4=
λ2-5λ=0
λ (λ-5) =0
λ = 0
λ = 5
At λ = 0 point, the corresponding vectors are
Matrix B =) =)
Solving vectors BX=O
Using row reduction
)( ) = ()
) =
Let R2 = R1 – 2R2
4x1 - 2x2 =0
x2= -2 x1
let x1 = 1
x2 = 2
Thus characteristic vector x1 =)
At λ = 5 point, the corresponding vectors are
C =) =)
CX = 0
Using low reduction
)( ) = ()
)
2R1 + R2 = R2
-X1 + 2X2 = 0
X1 = 2 X2
Let X2 = 1, X1 = 2
Thus
)( ) = ) = )
Characteristics vector X2 =)