ECON3014 Mathematical Economics Comprehensive Notes

ECON3014 Mathematical Economics builds the mathematical tools that economists use to formalize economic reasoning—optimisation, constrained choice, duality, equilibrium analysis, and comparative statics. These notes provide a rigorous, exam-focused pathway through the most commonly assessed concepts and methods, with worked examples tailored to typical university-level expectations at South African institutions (including the University of the Witwatersrand context within the Wits Economics Exam Notes collection).

1. Core Mathematical Foundations for ECON3014

Mathematical economics is not “extra math”; it is a translation layer between economic statements and mathematical objects (functions, sets, vectors, matrices, and operators). The exam in ECON3014 typically rewards students who can (i) set up the economic problem precisely, (ii) apply the right mathematical technique, and (iii) interpret the results economically.

1.1 Functions, Domains, and Economic Interpretation

A large portion of early exam questions can be solved if you are disciplined about definitions:

Variable naming and dimensions
- Scalars: (x, p)
- Vectors: (\mathbf{x}\in \mathbb{R}^n)
- Matrices: (A\in \mathbb{R}^{n\times n})
Domain matters
- Consumption variables might be restricted: (x\ge 0)
- Production inputs typically have constraints: (L\ge 0, K\ge 0)
- Utility often requires monotonicity: “more is better” implies increasing in goods.

A typical exam translation:

“Households consume two goods (x) and (y)” (\Rightarrow) a utility function (u(x,y))
“Budget constraint” (\Rightarrow) (px + qy \le I)
“Feasible set” (\Rightarrow) ({(x,y)\in \mathbb{R}^2_+ : px+qy\le I})

1.2 Limits, Continuity, and Why They Matter

You rarely need advanced real analysis proofs in ECON3014, but you often need to justify existence and continuity of solutions:

Continuity ensures that small changes in parameters (prices, income) cause small changes in objective values.
Concavity/convexity ensures global optimum uniqueness (for many standard cases).

In constrained optimisation, you’ll often use:

Weierstrass-type intuition: if the feasible set is closed and bounded and the objective is continuous, an optimum exists.
If the feasible set is not bounded (e.g., (x\to\infty) allowed), you must check whether the objective prevents “escaping to infinity” (e.g., utility bounded above for maximisation with concavity, or cost increasing for minimisation).

1.3 Differentiation: Gradients, Hessians, and Second-Order Conditions

A common exam pattern is: define an optimisation problem, compute first-order conditions (FOCs), then use second-order conditions (SOCs).

For a scalar function (f:\mathbb{R}^n\to \mathbb{R}):

The gradient is (\nabla f(\mathbf{x}) = \left(\frac{\partial f}{\partial x_1},\dots,\frac{\partial f}{\partial x_n}\right)).
The Hessian is (H_f(\mathbf{x}) = \left[\frac{\partial^2 f}{\partial x_i\partial x_j}\right]_{i,j}).

Second-order test (typical exam statement):

If maximising (f) with (H_f) negative definite on the feasible direction set, you have a local maximum.

Example forms:

For single-variable problems: if (f''(x^*)<0), local maximum.
For two variables: if Hessian is negative definite, maximum.

1.4 Convexity, Concavity, and Global Optimality

Convexity/concavity are central because they turn local logic into global guarantees.

A function (f) is concave if for all (\lambda\in[0,1]),
[
f(\lambda x+(1-\lambda)y)\ge \lambda f(x)+(1-\lambda)f(y).
]
A function (f) is convex if the inequality flips.

Economic implications:

Utility maximisation: if (u(\cdot)) is concave, utility maximisation under linear constraints yields a well-behaved solution.
Cost minimisation: if the cost function is convex in inputs, the minimisation problem is well-behaved.

Exam shortcut:

Many preference/technology problems use functional forms that are concave/convex by design (e.g., log utility, Cobb–Douglas utility, quadratic costs with positive coefficients).

1.5 Taylor Expansion and Local Comparative Statics

In comparative statics, Taylor expansions are often used implicitly.

For a differentiable function (f(x)) near (x_0):
[
f(x)\approx f(x_0)+f'(x_0)(x-x_0).
]

In multivariate versions:
[
f(\mathbf{x}) \approx f(\mathbf{x}_0)+\nabla f(\mathbf{x}_0)^\top(\mathbf{x}-\mathbf{x}_0).
]

Economically, this supports:

“Small change in price causes small change in demand”
Deriving elasticities and interpreting sensitivity using derivatives.

2. Optimisation and Duality (Consumer Choice, Producer Choice, and the Lagrangian)

This section covers the most exam-relevant machinery: constrained optimisation via Lagrange multipliers and the logic behind dual formulations.

2.1 Unconstrained Optimisation: Basic Templates

Template A: interior solution

Write objective ( \max_x f(x)).
FOC: (f'(x)=0).
SOC: (f''(x)<0) for a maximum.
Check boundaries if domain restricts.

Template B: quadratic objectives
If the objective has the form (-ax^2+bx+c) with (a>0), the maximum is at
[
x^*=\frac{b}{2a}
]
and the maximum value is obtained by substitution.

This template appears frequently in exam questions about cost minimisation or welfare analysis with quadratic forms.

2.2 Constrained Optimisation: Lagrange Multipliers

For maximising (f(x,y)) subject to (g(x,y)=0):
[
\max_{x,y} f(x,y)\quad \text{s.t.}\quad g(x,y)=0.
]
Construct the Lagrangian
[
\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda g(x,y).
]
Then solve:

( \frac{\partial \mathcal{L}}{\partial x}=0)
( \frac{\partial \mathcal{L}}{\partial y}=0)
( g(x,y)=0)

Economic interpretation of (\lambda):

In constrained maximisation with resource constraints, (\lambda) often measures the marginal value of relaxing the constraint.
In constrained minimisation, it often measures marginal savings.

Exam interpretation checklist

If objective is utility (u) and constraint is budget (px+qy=I), then (\lambda) can be linked to marginal utility of income.
If objective is cost (C) and constraint is producing a target output (q=\bar{q}), then (\lambda) relates to marginal cost of increasing output.

2.3 Inequality Constraints and Karush–Kuhn–Tucker (KKT) Conditions

Many exam problems include non-negativity constraints:
[
x\ge 0,\quad y\ge 0.
]
For problems with inequality constraints, Lagrange multipliers generalise via KKT conditions.

For constraint (h(x)\le 0), multipliers satisfy:

(\mu \ge 0)
Complementary slackness: (\mu, h(x)=0)

A standard example:
[
\max_{x\ge 0} u(x)\quad \text{s.t.}\quad px\le I.
]
You test:

Is the unconstrained optimum feasible?
If not, boundary solution emerges.

Boundary solution logic

If at optimum (x=0), then check whether the marginal utility of increasing (x) from zero is positive or the marginal utility is too low relative to price.

2.4 Consumer Theory: Budget Constraint, Demand, and Marshallian vs Hicksian

2.4.1 Marshallian (Uncompensated) Demand

Start with:

Utility (u(x,y))
Budget: (px+qy\le I)

Set up:
[
\max_{x,y\ge 0} u(x,y)\quad \text{s.t.}\quad px+qy\le I.
]
If optimum uses full income (typical with strictly increasing preferences), then:
[
px+qy=I.
]

From the solution ((x^(p,q,I),y^(p,q,I))), you obtain Marshallian demands:
[
x_M(p,q,I),\quad y_M(p,q,I).
]

2.4.2 Hicksian (Compensated) Demand and Expenditure Minimisation

Define the expenditure function:
[
e(p,q,\bar{u}) = \min_{x,y\ge 0} { px+qy : u(x,y)\ge \bar{u}}.
]
The Hicksian demands minimise expenditure required to reach utility (\bar{u}).

This dual approach is crucial because it leads to stronger mathematical properties:

Differentiability of the expenditure function under regularity conditions.
Relationships between Marshallian and Hicksian demands via compensation.

2.4.3 Duality Links (Exploitable Exam Results)

Key duality facts frequently used:

Envelope theorem links derivatives of value functions (e.g., expenditure) to Lagrange multipliers.
Under regularity conditions (e.g., local non-satiation, convexity of preferences), Marshallian and Hicksian demands are connected.

A frequent exam move:

Solve expenditure minimisation for Hicksian demands first (sometimes easier due to homogeneity and linear budgets).
Then use Shephard’s lemma:
[
\frac{\partial e(p,q,\bar{u})}{\partial p} = x_H(p,q,\bar{u}),\quad \frac{\partial e}{\partial q}= y_H(p,q,\bar{u}).
]

2.5 Worked Example: Cobb–Douglas Utility and Marshallian Demand

Consider utility:
[
u(x,y)=x^{\alpha}y^{1-\alpha},\quad 0<\alpha<1,
]
with budget (px+qy=I).

Maximise:
[
\max_{x,y\ge0} x^{\alpha}y^{1-\alpha}\quad \text{s.t.}\quad px+qy=I.
]

Because Cobb–Douglas is quasi-concave and interior for positive (\alpha), focus on interior solutions.

FOCs via Lagrangian:
[
\mathcal{L}=x^{\alpha}y^{1-\alpha}+\lambda(I-px-qy).
]

Partial derivatives:
[
\frac{\partial \mathcal{L}}{\partial x}=\alpha x^{\alpha-1}y^{1-\alpha}-\lambda p=0
]
[
\frac{\partial \mathcal{L}}{\partial y}=(1-\alpha)x^{\alpha}y^{-\alpha}-\lambda q=0.
]

Divide the two equations to eliminate (\lambda):
[
\frac{\alpha x^{\alpha-1}y^{1-\alpha}}{(1-\alpha)x^{\alpha}y^{-\alpha}}=\frac{p}{q}
]
Simplify:
[
\frac{\alpha}{1-\alpha}\cdot \frac{y}{x}=\frac{p}{q}
\quad\Rightarrow\quad
\frac{y}{x}=\frac{1-\alpha}{\alpha}\cdot \frac{p}{q}.
]
So:
[
y = x\cdot \frac{1-\alpha}{\alpha}\cdot \frac{p}{q}.
]
Plug into budget:
[
px + q\left(x\cdot \frac{1-\alpha}{\alpha}\cdot \frac{p}{q}\right)=I
\Rightarrow
px + x\cdot \frac{1-\alpha}{\alpha}p = I
\Rightarrow
px\left(1+\frac{1-\alpha}{\alpha}\right)=I.
]
Since (1+\frac{1-\alpha}{\alpha}=\frac{1}{\alpha}), we get:
[
px\cdot \frac{1}{\alpha} = I
\Rightarrow
x^=\alpha \frac{I}{p}.
]
Similarly:
[
y^=(1-\alpha)\frac{I}{q}.
]

Exam pay-off

You can show that the fraction of income spent on each good is constant:
[
px^=\alpha I,\quad qy^=(1-\alpha)I.
]
Demand functions are linear in income and inversely proportional to prices, implying constant expenditure shares.

2.6 Producer Theory: Cost Minimisation and Profit Maximisation

2.6.1 Cost Minimisation

Given production technology (q=f(K,L)) and input prices (r,w), define:
[
\min_{K,L\ge0} , rK+wL \quad \text{s.t.}\quad f(K,L)\ge \bar{q}.
]
The value function is the cost function (C(r,w,\bar{q})).

First-order conditions lead to tangency conditions between input isoquants and cost lines:

At optimum, the marginal rate of technical substitution equals the relative input price ratio.

2.6.2 Profit Maximisation

Profit is:
[
\pi = p q – C(r,w,q).
]
A common approach:

Maximise profit over output (q) given output price (p) and input prices summarised in (C(\cdot)).
FOC:
[
p = \frac{\partial C}{\partial q}(r,w,q)=MC.
]
SOC depends on convexity of cost.

2.7 Duality in Production: Shephard’s Lemma for the Cost Function

If cost function exists and is differentiable:

Shephard’s lemma:
[
\frac{\partial C(r,w,q)}{\partial r} = K^(r,w,q),\quad \frac{\partial C}{\partial w}=L^(r,w,q).
]
This tells you that derivatives of the cost function with respect to input prices give the conditional factor demands.

This is a common exam trick: instead of re-solving the optimisation, use properties of the value function.

3. Equilibrium Models, Stability, and Comparative Statics

Equilibrium analysis in mathematical economics is typically taught through systems of equations and functions, then examined through comparative statics (how equilibrium changes when parameters change). This section formalises equilibrium and builds the comparative-statics toolkit.

3.1 General Equilibrium as Fixed Points

Many equilibrium problems can be written as fixed points.

Suppose market demand and supply for a good are:
[
D(p,\theta),\quad S(p,\theta),
]
where (\theta) is a parameter (income, technology, preferences).
Equilibrium price (p^) solves:
[
D(p^,\theta)=S(p^,\theta).
]
Define excess demand:
[
Z(p,\theta)=D(p,\theta)-S(p,\theta).
]
Equilibrium is:
[
Z(p^,\theta)=0.
]

This formulation is powerful because it connects equilibrium with root-finding and implicit function results.

3.2 Implicit Function Theorem: Comparative Statics Backbone

If equilibrium satisfies:
[
Z(p^,\theta)=0,
]
and ( \frac{\partial Z}{\partial p}(p^,\theta)\neq 0), then locally there exists a differentiable function (p(\theta)) such that (Z(p(\theta),\theta)=0).

Differentiate implicitly:
[
\frac{d}{d\theta}Z(p(\theta),\theta)=Z_p \cdot \frac{dp}{d\theta}+Z_\theta=0.
]
So:
[
\frac{dp}{d\theta} = -\frac{Z_\theta}{Z_p}.
]

Exam interpretation

Determine signs:
- Usually (Z_p<0) if demand decreases with price and supply increases with price (typical).
If (Z_\theta>0) (a parameter increases excess demand), then (dp/d\theta>0).

This sign analysis is a major part of comparative statics questions.

3.3 One-Good Partial Equilibrium Comparative Statics

Assume:

(D_p<0)
(S_p>0)
So:
[
Z_p = D_p – S_p <0.
]
Now consider a parameter (\theta) that shifts demand. If:
(D_\theta>0) and (S_\theta=0), then:
[
Z_\theta = D_\theta – 0 >0.
]
Thus:
[
\frac{dp}{d\theta}=-\frac{Z_\theta}{Z_p} >0.
]
So equilibrium price rises.

Worked sign example

If income increases and the good is normal, demand rises:
- (D_I>0), (S_I=0)
- Price increases.

3.4 Multimarket Comparative Statics and Linearisation

For multiple goods, equilibrium becomes a system:
[
\mathbf{Z}(\mathbf{p},\boldsymbol{\theta})=\mathbf{0},
]
where (\mathbf{p}\in\mathbb{R}^n).

Linearising near equilibrium:

Use Jacobians:
[
J = \frac{\partial \mathbf{Z}}{\partial \mathbf{p}}
]
and parameter derivatives:
[
\frac{\partial \mathbf{Z}}{\partial \boldsymbol{\theta}}.
]
Then:
[
J\cdot d\mathbf{p} + \frac{\partial \mathbf{Z}}{\partial \boldsymbol{\theta}}d\boldsymbol{\theta} = 0
\Rightarrow
d\mathbf{p}= -J^{-1}\frac{\partial \mathbf{Z}}{\partial \boldsymbol{\theta}}d\boldsymbol{\theta}.
]

Exam expectation

You may be asked to compute (J^{-1}) for small systems (2 goods) or at least interpret signs.
Stability is separate but often appears in the same questions.

3.5 Stability: Adjustments Toward Equilibrium

Stability addresses whether an economy converges to equilibrium under a dynamic adjustment rule.

A simple tatonnement mechanism:

Price adjusts in proportion to excess demand:
[
\dot{p}= \phi Z(p,\theta),
]
where (\phi>0).

At equilibrium (p^), (Z(p^,\theta)=0). Consider small deviation (e=p-p^). Linearise:
[
\dot{e} \approx \phi Z_p(p^) e.
]
So stability requires:

If (\dot{e} = \phi Z_p e),
and (Z_p<0), then (e) decays (stable).

Thus for a one-good market with downward-sloping demand and upward-sloping supply, tatonnement is typically stable under standard assumptions.

3.6 Elasticities and Welfare-Relevant Responses

Comparative statics can be strengthened using elasticities.

For a good with demand (D(p)) and supply (S(p)), you might express response in relative terms:

Price elasticity of demand:
[
\varepsilon_D = \frac{dD}{dp}\cdot \frac{p}{D}.
]
Price elasticity of supply:
[
\varepsilon_S = \frac{dS}{dp}\cdot \frac{p}{S}.
]

In multi-step exam problems:

Use algebra to find equilibrium (p^*) from (D(p)=S(p)).
Compute how (p^*) changes with a shift parameter.
Translate to elasticities and interpret impacts on quantities and revenue.

3.7 Example: Equilibrium with Linear Demand and Supply

Let:
[
D(p)=a-bp,\quad S(p)=c+dp,
]
with (a,b,c,d>0).

Equilibrium:
[
a-bp = c+dp \Rightarrow a-c = (b+d)p \Rightarrow p^=\frac{a-c}{b+d}.
]
Quantity:
[
q^ = D(p^*) = a-b\frac{a-c}{b+d}.
]

Now consider comparative statics:

If (a) increases by (\Delta a), then:
[
\frac{\partial p^*}{\partial a}=\frac{1}{b+d}>0,
]
so price rises.
If (c) increases by (\Delta c), then:
[
\frac{\partial p^*}{\partial c}=-\frac{1}{b+d}<0,
]
so price falls because supply shifts up.

This is the core sign logic that repeats in many exam questions.

4. Dynamics, Linear Algebra Tools, and Market Design Under Mathematical Constraints

This section adds depth: matrix tools for systems, dynamic optimisation intuition, and constrained equilibrium problems. ECON3014 often expects you to manipulate expressions cleanly, compute determinants/inverses for small matrices, and interpret dynamic outcomes.

4.1 Linear Algebra Essentials for Economics

4.1.1 Vectors and Dot Products

For vectors (\mathbf{x},\mathbf{y}\in\mathbb{R}^n):
[
\mathbf{x}^\top \mathbf{y} = \sum_{i=1}^n x_i y_i.
]
Economic interpretation:

Dot products appear in budget constraints and in welfare functions (e.g., weighted sums).

4.1.2 Matrices, Rank, and Invertibility

For equilibrium systems:

Invertibility of Jacobians matters. In (d\mathbf{p} = -J^{-1}(\cdots)),
You need (\det(J)\neq 0).

For (2\times 2) matrices:
If
[
J=\begin{pmatrix}
a & b\
c & d
\end{pmatrix},
]
then
[
\det(J)=ad-bc,
]
and
[
J^{-1}=\frac{1}{ad-bc}\begin{pmatrix}
d & -b\
-c & a
\end{pmatrix}.
]

Exam questions often involve:

computing (J^{-1}),
multiplying by a shift vector,
interpreting sign patterns.

4.2 Dynamic Optimisation vs Dynamic Stability

Two different ideas can appear:

Dynamic optimisation: a decision-maker chooses an action path over time to maximise a dynamic objective (e.g., present-value utility).
Dynamic stability: an economy or system converges over time under an adjustment rule.

This section focuses mainly on stability and adjustment, but it also prepares you for dynamic optimisation structures (discount factors, intertemporal choice).

4.3 A Simple Discrete-Time Stability Model

Let a variable (x_t) follow:
[
x_{t+1}=g(x_t).
]
An equilibrium is (x^) such that (x^=g(x^*)).

Linearise around (x^): let (e_t=x_t-x^). Then:
[
e_{t+1}\approx g'(x^*) e_t.
]
Stability requires:

(|g'(x^*)|<1).

Economic meaning

If the adjustment rule overshoots too much ((|g'|\ge 1)), the system oscillates or diverges.

4.4 Intertemporal Choice: Present Value and Discounting (Foundational)

A common “math economics core” topic: maximising discounted utility:
[
\max \sum_{t=0}^{\infty} \beta^t u(c_t),
]
where (0<\beta<1) is the discount factor.

Even if ECON3014 doesn’t demand full Euler equation derivations, exam questions often expect:

correct handling of present value
algebraic manipulation of infinite series when utility or returns are linearised.

Infinite geometric series:
If (|r|<1),
[
\sum_{t=0}^{\infty} r^t = \frac{1}{1-r}.
]

A typical exam application:

Compute present value of constant cash flow (A) under discount factor (\beta):
[
PV=\sum_{t=0}^{\infty}\beta^t A = \frac{A}{1-\beta}.
]

4.5 Linear Quadratic Structure: Why It’s Exam-Friendly

Quadratic forms lead to tractable solutions. A standard structure:
[
\min_x \left( \frac{1}{2}x^\top A x – b^\top x \right),
]
with (A) positive definite.

FOC:
[
Ax=b \Rightarrow x^* = A^{-1}b.
]

This is a clean algebraic pattern many instructors use to test:

matrix derivatives,
definiteness assumptions,
inversion.

4.6 Constraints in Equilibrium: Feasible Sets and Projections

In equilibrium or optimisation, feasibility constraints define sets:
[
\mathcal{F}={\mathbf{x}: \mathbf{x}\ge 0,; B\mathbf{x}\le \mathbf{y}}.
]
Sometimes exam problems ask you to interpret “projection” onto feasible sets:

If an unconstrained optimum violates constraints, the solution lies on the boundary.
KKT conditions identify which constraints bind.

A typical two-constraint exam scenario:

maximise (u(x)) with (x\ge 0) and (ax\le I).
Determine whether (x^*=x_{\text{unconstrained}}) satisfies both.
If not, solve boundary cases:
- (x=0) or (ax=I).

4.7 Worked Example: Two-Market Comparative Statics with a Jacobian

Suppose excess demand system:
[
Z_1(p_1,p_2)=a_1 – b_1 p_1 + \gamma p_2,
]
[
Z_2(p_1,p_2)=a_2 – b_2 p_2 + \delta p_1,
]
with parameters (a_1,a_2) shifting demand, and (b_1,b_2>0). Equilibrium solves:
[
Z_1=0,\quad Z_2=0.
]

Compute Jacobian:
[
J=\frac{\partial (Z_1,Z_2)}{\partial (p_1,p_2)}=
\begin{pmatrix}
-b_1 & \gamma\
\delta & -b_2
\end{pmatrix}.
]
Then:
[
\det(J)=(-b_1)(-b_2)-\gamma\delta=b_1 b_2-\gamma\delta.
]
If (b_1 b_2-\gamma\delta\neq 0), inverse exists.

Now if both (a_1) and (a_2) shift by a small amount (d a_1, d a_2), parameter vector derivative is:
[
\frac{\partial \mathbf{Z}}{\partial \mathbf{a}}=
\begin{pmatrix}
1 & 0\
0 & 1
\end{pmatrix}.
]
Thus:
[
d\mathbf{p} = -J^{-1} , d\mathbf{a}.
]

Even without computing full expressions, exam questions often ask for signs:

If cross effects (\gamma,\delta>0), an increase in one demand parameter may raise both prices.
But exact sign requires multiplication and determinant sign.

This example illustrates how linear algebra operationalises comparative statics.

5. Exam-Ready Problem Solving: Templates, Common Pitfalls, and South African University-Style Practice Cases

This final section consolidates everything into exam strategies, with worked mini-cases and a checklist for typical marking schemes. It also focuses on how to present solutions clearly—often the difference between partial and full marks.

5.1 How ECON3014 Examiners Allocate Marks (Implicit Logic)

While mark schemes vary, typical allocation follows a predictable structure:

Correct set-up (30–40%)
- Identify objective and constraints.
- Use correct notation and specify domains.
Correct method (30–40%)
- Use Lagrangian/KKT correctly.
- Apply implicit differentiation correctly for comparative statics.
Correct algebra (20–30%)
- Solve for decision variables.
- Compute derivatives and simplify without errors.
Economic interpretation (10–20%)
- Explain sign of comparative statics.
- Interpret multipliers and equilibrium adjustments.

So, even if your final expression is close but not exact, showing correct set-up and method can still earn substantial marks.

5.2 Presentation Templates That Win Marks

5.2.1 Optimisation Answer Structure

Use this pattern:

Problem statement
- “Maximise (u(x,y)) subject to (px+qy\le I).”
Lagrangian
- Write (\mathcal{L} = u(x,y)+\lambda(I-px-qy)).
FOCs and constraint
- Show partial derivatives step-by-step.
Solve system
- Substitute or divide to eliminate (\lambda).
Second-order check
- State concavity/convexity or show sign of Hessian along relevant directions.
Economic meaning
- Interpret multipliers and verify budget binding.

5.2.2 Comparative Statics Answer Structure

Use:

Define equilibrium condition
- (Z(p,\theta)=D(p,\theta)-S(p,\theta)=0)
Compute derivatives
- (Z_p, Z_\theta)
Implicit differentiation
- (\frac{dp}{d\theta}=-\frac{Z_\theta}{Z_p})
Sign argument
- Determine sign using assumptions on monotonicity.
Interpretation
- “As income rises, demand increases so excess demand shifts up; price increases.”

5.3 Common Pitfalls (and How to Avoid Them)

Forgetting interior vs boundary
- If (x\ge 0) and the solution from FOCs implies (x<0), you must switch to boundary solution (KKT logic).
Mismatched inequality direction
- In KKT, remember constraint form:
  - For (h(x)\le 0), multiplier (\mu\ge 0).
  - Signs matter.
Confusing Marshallian and Hicksian
- Marshallian solves utility maximisation under budget.
- Hicksian solves expenditure minimisation under utility target.
Incorrect derivative of products
- Students often slip in Cobb–Douglas or log objectives.
- Use log transformations carefully:
  - If (u=x^\alpha y^{1-\alpha}), maximising (u) is equivalent to maximising (\ln u = \alpha\ln x + (1-\alpha)\ln y) for interior solutions.
Comparative statics sign errors
- Always compute (Z_p) correctly:
  - If (Z=D-S), then (Z_p = D_p – S_p).
- Don’t accidentally use (D_p + S_p).

5.4 Practice Case Studies (Worked Mini-Problems)

Case Study A: Single Consumer, Log Utility, Two Goods

Let utility:
[
u(x,y)=\ln x + \beta \ln y,\quad \beta>0.
]
Budget (px+qy=I). Solve:
[
\max_{x,y>0} \ln x + \beta \ln y \quad \text{s.t.}\quad px+qy=I.
]

Lagrangian:
[
\mathcal{L}=\ln x + \beta \ln y + \lambda (I-px-qy).
]
FOCs:
[
\frac{1}{x}-\lambda p=0 \Rightarrow \lambda=\frac{1}{px}
]
[
\frac{\beta}{y}-\lambda q=0 \Rightarrow \lambda=\frac{\beta}{qy}.
]
Set equal:
[
\frac{1}{px}=\frac{\beta}{qy}\Rightarrow y=\beta \frac{p}{q}x.
]
Budget:
[
px+q\left(\beta \frac{p}{q}x\right)=I
\Rightarrow
px+\beta p x = I
\Rightarrow
p x(1+\beta)=I
\Rightarrow
x^=\frac{I}{p(1+\beta)}.
]
Then:
[
y^=\beta \frac{p}{q}x^*=\beta \frac{p}{q}\cdot \frac{I}{p(1+\beta)}=\frac{\beta I}{q(1+\beta)}.
]

Economic interpretation

Expenditure shares are constant:
[
px^=\frac{I}{1+\beta},\quad qy^=\frac{\beta I}{1+\beta}.
]

This mirrors the Cobb–Douglas result and is a common mark-friendly pattern.

Case Study B: Producer with Quadratic Cost

Suppose a firm chooses output (q\ge 0). Profit:
[
\pi(q)=p q – \left(\frac{1}{2}a q^2 + b q\right),
]
with (a>0).

Profit maximisation:
[
\max_{q\ge 0} , p q – \frac{1}{2}a q^2 – b q
= \max_{q\ge 0} \left( (p-b)q – \frac{1}{2}a q^2 \right).
]
FOC:
[
(p-b) – a q = 0 \Rightarrow q^*=\frac{p-b}{a}.
]
Boundary check:

If (p-b<0), then (q^<0) so constraint implies (q^=0).
So:
[
q^*=\max\left{0,\frac{p-b}{a}\right}.
]

Exam interpretation

Output increases linearly in price above the shutdown threshold (p=b).
This type of “piecewise solution” is a typical test of constraint awareness.

Case Study C: Comparative Statics with Excess Demand Shifts

Consider:
[
D(p,\theta)=\theta – p,\quad S(p)=p-1.
]
Equilibrium solves:
[
\theta – p = p – 1 \Rightarrow \theta + 1 = 2p \Rightarrow p^=\frac{\theta+1}{2}.
]
Thus:
[
\frac{dp^}{d\theta}=\frac{1}{2}>0.
]
Quantity:
[
q^=D(p^,\theta)=\theta – \frac{\theta+1}{2} = \frac{\theta-1}{2}.
]
So:
[
\frac{dq^*}{d\theta}=\frac{1}{2}.
]

Interpretation

As (\theta) increases (demand shifts right), both equilibrium price and quantity rise.
If (\theta<1), quantity becomes negative in this linear model; an exam might require you to consider non-negativity constraints (q\ge 0) which would change the feasible equilibrium to a boundary case (e.g., (q=0)). Being alert to feasibility yields higher marks.

5.5 South African Context: University Exam Style Considerations (Wits Economics Notes Framing)

Within South African university economics teaching and assessments, mathematical economics papers often emphasise:

clarity of derivation (a clean sequence of steps),
correct handling of assumptions (monotonicity/concavity),
and correct interpretation of results.

For students studying at institutions such as the University of the Witwatersrand (Wits) (within the Wits Economics Exam Notes collection), it is especially important to:

present the economic model before heavy algebra,
explicitly state when assumptions imply interior solutions,
and when you use a theorem (e.g., implicit function theorem or envelope theorem), you should explain the minimal conditions needed (e.g., differentiability, non-zero Jacobian determinant).

Even if a specific exam question does not mention the theorem by name, your answer should mirror its logic.

5.6 Final Master Checklist for ECON3014

Before submitting an exam solution, verify:

Optimisation

Objective and constraints are written correctly.
Domain constraints considered (non-negativity, feasibility).
Lagrangian sign conventions are correct.
FOCs derived without missing terms.
Solution checked for interior/boundary.
Concavity/convexity used for global logic when applicable.
Multiplier interpreted economically.

Comparative Statics

Equilibrium condition written as an equation system.
Excess demand (or equilibrium gap) defined consistently.
Correct derivative signs: (Z_p) and (Z_\theta).
Implicit differentiation applied correctly.
Feasibility constraints not violated.
Final sign statement is justified.

Linear Algebra / Multi-market

Jacobian correctly computed.
Determinant non-zero if inversion is used.
Matrix multiplication dimensionally consistent.
Sign interpretation matches algebraic results.

Summary of Key Takeaways

Optimisation: Lagrangians, KKT, and second-order conditions convert economic problems into solvable systems.
Duality: expenditure and cost functions plus envelope/Shephard’s lemma link derivatives to demand and factor choices.
Equilibrium and comparative statics: fixed-point and implicit-function thinking yields sign and sensitivity results efficiently.
Dynamics and stability: linearisation around equilibria clarifies whether adjustment rules converge.
Linear algebra: Jacobians and matrix inversion operationalise multi-variable comparative statics.

These tools form a coherent foundation for ECON3014. Mastery comes from repeatedly moving between economic statements and mathematical objects—then interpreting the math back into economic meaning.