Economics: Mathematics for Economics Assignment Solution Analysis

This document presents a detailed solution to a mathematics for economics assignment. The solution covers several key economic concepts, including the calculation of consumer and producer surplus using demand and supply functions, and the determination of market equilibrium. It explores capital formation using integration, and analyzes the time required for capital stock to exceed a certain value. Additionally, the assignment delves into the stability of market models and two-sector models using differential and difference equations. The solutions include step-by-step calculations, explanations, and graphical representations where relevant, providing a comprehensive understanding of the economic principles involved. The assignment also discusses the stability of the market and provides solutions to various economic models. The document includes references to relevant economic literature.

Running head: MATHEMATICS FOR ECONOMICS 1
Mathematics for Economics
Firstname Lastname
Name of Instituion

MATHEMATICS FOR ECONOMICS 2
Question 2a
Given the demand function P = Q2D – 4QD + 68 and supply function P = Q2s + 2Qs +12.
Assuming pure competition, find
The consumer surplus
The producer surplus
solution
At equilibrium, the supply price is equal to demand price, therefore we equate the two equations
as follows
- Q2 - 4Q + 68 = Q2 - 2Q + 12
Collecting the like terms, we have 2Q2 + 2Q - 56 = 0
The quadratic equation is formed and solved, Q2 + Q - 28 = 0
The solution gives the value of Q to be Q = 5 and -6, hence we consider Q=5 since Q >= 0
Price, P = (5 x 5) - (2 x 5) + 12 = 25 - 10 + 12 = 27
(a) Using the demand function, we can solve the value for the consumer surplus
Taking the value of QD to be = 0, the price value P = 68
Value of Consumer surplus is determined by getting the area between demand curve and market
price
= (1/2) x 5 x (68 - 27)
= (1/2) x 5 x 41 = 102.5
(b) Producer surplus is determined by applying the supply equation.
By replacing the value of QS in the equation by 0, the minimum acceptable price P becomes,
P=12
The value of producer surplus is equal to the area between supply curve and market price
= (1/2) x 5x (27 - 12) = (1/2) x 5 x15 = 37.5
Question 2 b
If the net investment function is given by I(t) = 100e0.1t calculate

MATHEMATICS FOR ECONOMICS 3
i. The capital formation from the end of the second year to the end of the fifth year
Solution
The required total capital information ¿∫
2
5
100 e0.1 t dt
100(10√ e −105√ e )
Simplify/rewrite:
1000(√ e −5√ e )
The capital formation becomes = 427.32
ii. The number of years required before the capital stock exceed 100000
Solution
In this case we equate the value 100000 to the intrgral and make the constant to be 0

MATHEMATICS FOR ECONOMICS 4
1000e(t/10) > 100000
Hence e(t/10) > 100
Solving for t, we (t/10) > 4.60
Time = 46 years
c. Write down an expression for g’(x) when g(x) = ( 5t2 -2t)dt
Question 2 a
The producer's surplus at Q=a for the supply function P = 6 + 8Q is known to be 400. Find
the value a
Solution
The producer surplus Q =a for supply equation of P = S(Q) = 6 + 8Q but PS(a) = 400, the value
of a need to be determined

MATHEMATICS FOR ECONOMICS 5
Here Q1 = a and P1 = 5(a) = 6 + 8 a
Hence using the above equation PS (Q) = P1Q1 - ∫
0
Q 1
S ( Q ) dQ
We thereafter have 400 = (6 +8a) a - ∫
0
a
(6+8 Q)dQ
400 = 6a +8a^2 – (6a +4a^2) = 6a + 8a^2 -6a – 4a^2
400 = 4a^2 resulting into a^2 = 100 hence a = 10
Thus the value of a = 10
Question 2 b
The demand function is given by P = 74 -Q2D and the supply function P = (Qs +2)2.
Calculate the consumer's and producer's surplus under pure competition

MATHEMATICS FOR ECONOMICS 6
Question 3
a) Consider the supply and demand equations:
Solution
Qst = Pt-1 -8
QDt = -2Pt + 22
At equilibrium, Pt-1 -8 = -2Pt + 22
By collecting the terms, 3pt = 30
Solving for Pt , Pt = 30/3 = 10

MATHEMATICS FOR ECONOMICS 7
Substituting values Q = -2Pt + 22 = -2(10) +22 = 2
The time path of the price is given by, Pt = Apt(0) where A is the slope
A = slope of supply function
slope of demand function
= 1/-2 = -0.5
Since the value of A is less than 1, oscillation is damped therefore the market is stable.
b) consider the two-sector model
dy/dt = 0.1 (C+I-Y)
C= 0.9Y +100
I=300
Find an expression of Y(t) when Y (0) = 2000. is this system stable or unstable?
Substituting the expressions for C and I into
Dy/dt = 0.1(C + I − Y )
gives
dy/dt = 0.1(0.9Y + 100 + 300 − Y)
= −0.01Y + 40
The complementary function is given by
CF = Ae−0.1t
and for a particular solution we try
Y(t) = D
for some constant, D. Substituting this into the differential equation gives
0 = −0.01D + 40
which has solution D = 4000. The general solution is therefore
Y(t) = Ae−0.01t + 4000
The initial condition
Y(0) = 2000
gives
A + 4000 = 2000
and so A is -3000. The solution is

MATHEMATICS FOR ECONOMICS 8
Y(t) = -3000e−0.01t + 4000
The first term is a negative exponential, so it converges to zero as t increases. Consequently, Y(t)
eventually settles down to an equilibrium value of 4000 and the system is stable.
Question 3
a) consider the two -sector model: Yt = Ct + It
Ct = 0.9Yt-1 + 250
It = 350
solution
Substituting the expressions for Ct and It into
Yt = Ct + It
gives
Yt = (0.9Yt−1 + 250) + 350
= 0.9Yt−1 + 600
The complementary function is given by
CF = A (0.9) t
and for a particular solution we try
Yt = D
for some constant D. Substituting this into the difference equation gives
D = 0.9D + 600
which has solution D = 6000. The general solution is therefore
Yt = A (0.9) t + 6000
The initial condition,
Y0 = 6500
gives
6500 = A (0.9)0 + 6000 = A + 6000
and so, A is 500. The solution becomes
Yt = 500(0.9) t + 6000
As t increases, (0.9) t converges to zero and so Yt eventually settles down at the equilibrium level
of 6000. The system is therefore stable.
a) consider the market model

MATHEMATICS FOR ECONOMICS 9
Qs = 2P-2
QD = -P + 4
dp/dt= 1/3 (QD - Qs)
Finding the expression for P(t), Qs(t) and QD(t) and stating whether the system is stable or
unstable
Solution
For equilibrium we equate the two expression
2P-2 = -P + 4
3P = 6 hence P = 2
Substituting the value of P to get Q, Q = 2P – 2 = (2 x 6) -2 = 10
Time path of price, P(t) = APt(0)
Where A = slope of supply function
slope of demand function
A = -1/2 = -0.5 hence the system is stable
Qst = 2Pt -2
QDt = -Pt-1 +4
dp/dt= 1/3 (QD - Qs) = 1/3(-P+4 -2P +2) = 1/3( -3P-6)
finding P(t), 1 = 1/3(-Pt-1 +4 - 2Pt -2) = 1/3( -3P(t-1) + 6)
1 = -Pt-1 + 2
P(t) = 1
Qst = 2Pt -2
QDt = -Pt-1 +4
References
Buchanan, J. M. (2001). Game theory, mathematics, and economics. Journal of Economic
Methodology, 8(1), 27-32.
Leontief, W. (2015). Mathematics in economics. Bulletin of the American Mathematical Society,
60(3), 215-233.
Hoy, M., Livernois, J., McKenna, C., Rees, R., & Stengos, T. (2011). Mathematics for
economics. MIT press.
Ramsey, F. P., Mellor, D. H., & Stone, R. (2009). Foundations, Essays in Philosophy, Logic,
Mathematics and Economics.

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