INTERMEDIATE MICROECONOMICS
EXERCISE SOLUTIONS MANUAL
KOEN HANEGREEFS
VUB
,Table of Contents
About this Solutions Manual ..................................................................................................................... 1
Chapter 1 — Budget Constraint, Preferences & Utility (H2, H3, H4)............................................................. 2
Solutions — WPO 1 .............................................................................................................................. 2
Chapter 2 — Consumer Choice & Individual Demand (H5, H6) ................................................................... 9
Solutions — WPO 2 .............................................................................................................................. 9
Chapter 3 — Slutsky Equation & Market Demand (H8, H15)...................................................................... 11
Solutions — WPO 3 ............................................................................................................................ 11
Chapter 4 — Intertemporal Choice & Uncertainty (H10, H12) ................................................................... 22
Solutions — WPO 4 ............................................................................................................................ 22
Chapter 5 — Technology & Profit Maximisation (H19, H20) ....................................................................... 26
Solutions — WPO 5 ............................................................................................................................ 26
Chapter 6 — Cost Minimisation & Cost Curves (H21, H22) ....................................................................... 30
Solutions — WPO 6 ............................................................................................................................ 30
Chapter 7 — Firm Supply & Industry Supply (H23, H24) ............................................................................ 36
Solutions — WPO 7 ............................................................................................................................ 36
Chapter 9 — Monopoly & Monopoly Behaviour (H25, H26) ....................................................................... 41
Solutions — WPO 9 ............................................................................................................................ 41
Chapter 10 — Oligopoly & Game Theory (H28, H29, H30) ......................................................................... 49
Solutions — WPO 10 ........................................................................................................................... 49
About this Solutions Manual
This file contains the full worked solutions to every WPO exercise for chapters 1 to 10 of Intermediate
Microeconomics (VUB, Prof. Nikolas Vander Vennet). Use it in conjunction with the Study Document.
1
,Chapter 1 — Budget Constraint, Preferences & Utility (H2, H3, H4)
Solutions — WPO 1
Course: Intermediate Microeconomics (Varian) — VUB
Source: Exercises_1 H2, 3, 4.pdf
Instructor: Atanas Djondjourov
Chapter 2: Budget Constraint
Exercise 2.1
Solution:
Input: x₁ = 100 units; x₂ = 50 units; p₁ = €2; p₁’ = €3; p₂ = €4
Budget constraint: m = p₁ · x₁ + p₂ · x₂
• Initial prices: m = 2 · 100 + 50 · 4 = €400
• New prices: m = 3 · 100 + 50 · 4 = €500
Answer: Jan’s income should increase by €100.
Exercise 2.2
Solution:
Input: x₁ = 8, x₂ = 8; x₁’ = 10, x₂’ = 4; p₁ = €0.50
Budget constraint: m = p₁ · x₁ + p₂ · x₂
Both bundles (8, 8) and (10, 4) are on the budget line: - m = 8 · 0.5 + 8 · p₂ = 6 - m = 10 · 0.5 + 4 · p₂
Solving: m = 6; p₂ = 0.25
Budget line equation: 6 = 0.5x₁ + 0.25x₂
Intercepts: - x₁ = 6/0.5 = 12 - x₂ = 6/0.25 = 24
Answer: Amy gets €6 of pocket money per week.
2
,Exercise 2.3
Solution A:
Input: m = 60,000; p₁ (x₁ ≤ 1,000 m³) = €15; p₁ (x₁ > 1,000 m³) = €10; p₂ = €10
Budget constraints: - If x₁ ≤ 1,000: 60,000 = 15x₁ + 10x₂ - If x₁ > 1,000: 60,000 = 15 · 1,000 + 10(x₁ - 1,000) +
10x₂
So: 55,000 = 10x₁ + 10x₂
Key points: - When x₁ = 0: x₂ = 6,000 - When x₁ = 1,000: x₂ = 4,500 (kink point) - When x₂ = 0: x₁ = 5,500
Answer: The budget line has a kink at (1,000, 4,500).
Solution B:
Input: Same as (a) but with €6,000 connection costs
Budget constraints: - If x₁ ≤ 1,000: 60,000 = 6,000 + 15x₁ + 10x₂
So: 54,000 = 15x₁ + 10x₂ - If x₁ > 1,000: 60,000 = 6,000 + 15 · 1,000 + 10(x₁ - 1,000) + 10x₂
So: 49,000 = 10x₁ + 10x₂
Key points: - When x₁ = 0: x₂ = 5,400 - When x₂ = 0: x₁ = 4,900
Answer: The budget line shifts down (parallel shift) due to the fixed connection cost.
Exercise 2.4
Solution 1: Unconditional lump-sum subsidy
Input: m = 60,000; subsidy = €10,000
Budget constraint: 60,000 + 10,000 = C + O
So: 70,000 = C + O
Answer: The budget line shifts outward uniformly. Maximum C = 70,000 or maximum O = 70,000.
Solution 2: Conditional subsidy (minimum purchase requirement)
Input: m = 60,000; subsidy = €10,000 if C ≥ €30,000
Budget constraints: - When C < €30,000: 60,000 = C + O - When C ≥ €30,000: 70,000 = C + O
Key points: - Kink point at C = 30,000, O = 30,000 - Maximum C without subsidy = 60,000 - Maximum C with
subsidy = 70,000
Answer: Non-convex budget set due to discontinuity at C = €30,000.
3
,Solution 3: Proportional price subsidy (50% discount on computers)
Input: m = 60,000; subsidy = 50% on computers
Budget constraint: 60,000 = 0.5C + O
Key points: - Maximum O = 60,000 - Maximum C = 120,000
Answer: Effective price of computers falls to €0.50. Budget line becomes flatter.
Solution 4: Price subsidy with cap (50% discount, max €10,000 subsidy)
Input: m = 60,000; subsidy = 50% on computers (max = €10,000)
Transition point: Subsidy cap reached at C = €20,000 (subsidy = 0.5 × 20,000 = €10,000)
Budget constraints: - When C ≤ €20,000: 60,000 = 0.5C + O - When C > €20,000: 60,000 = 0.5 · 20,000 + (C -
20,000) + O
So: 70,000 = C + O
Key points: - Kink at C = 20,000, O = 50,000 - Maximum C = 70,000 - Slope changes from -0.5 to -1 at kink
Answer: Budget line has a kink at C = €20,000.
Chapter 3: Preferences
Exercise 3.1
Solution A:
p_lunch at 12h = €5; m = €15
m - 5 = €10
Answer: Ralph has €10 left.
Solution B:
p_lunch at 14h = 5 - 2 = €3
m - 3 = €12
Answer: Ralph has €12 left.
Solution C:
Budget line: Free budget = 15 - (5 - t) = 10 + t, where t is hours from noon
4
,The budget constraint is linear in time. Maximum free budget is at t = ±∞, but practically: - At 11h (1 hour
before): Free budget = 10 + 1 = €11 - At 12h (noon): Free budget = 10 - At 14h (2 hours after): Free budget = 10
+ 2 = €12
Indifference curve: At preferred lunch time of 11h, Ralph’s indifference curve is tangent to the budget line.
Exercise 3.2
Solution A:
Input: MRS (C > G) = -1/2; MRS (G > C) = -2
When G > C: MRS = -2, so Δx₂/Δx₁ = -2
For one extra cherry (Δx₁ = +1): Δx₂ = -2
Answer: Mary is willing to forgo 2 grapes for an extra cherry.
Solution B:
Input: MRS (C > G) = -1/2; MRS (G > C) = -2; initial bundle (11, 23)
The kink occurs where C = G (where the MRS switches from -2 to -1/2).
Starting from (11, 23) and moving toward (19, 23): - From (11, 23) to the kink: MRS = -2 - The kink is at (c, c)
where: -2 = (c - 23)/(c - 11) - Solving: -2(c - 11) = c - 23
-2c + 22 = c - 23
c = 15
So the kink is at (15, 15).
From (15, 15) to (19, ?): MRS = -1/2 - -1/2 = (x₂ - 15)/(19 - 15) - -1/2 · 4 = x₂ - 15 - x₂ = 13
Answer: Mary is indifferent between bundles (11, 23) and (19, 13).
Exercise 3.3
Solution:
Input: x₂ = constant - 4x₁^(1/2)
MRS calculation:
𝑑𝑥! 1 #"/! #"/!
MRS = = −4 ⋅ 𝑥" = −2𝑥"
𝑑𝑥" 2
At bundle (1, 6):
MRS = −2 ⋅ 1#"/! = −2
5
,Answer: The slope is -2. This is typical for quasilinear preferences, where MRS depends only on x₁.
Exercise 3.4
Solution:
A) Perfect substitutes — Both goods, “more is better”
Indifference curves are straight lines with negative slope (could be -1 if he truly doesn’t care about which
gender).
B) Both goods are bads — Satiation at origin
Indifference curves surround the origin; Freddy prefers bundles closer to (0, 0).
C) Perfect complements — Equal numbers required
Indifference curves are L-shaped, with the kink along the line men = women. Freddy needs equal numbers
to be happy.
D) One good, one bad — Men are good, women are bad
Indifference curves are horizontal lines. Freddy wants as many men as possible and as few women as
possible. Different numbers of men make him happy; women don’t matter (up to a limit—more women
make him worse off).
E) One good, one neutral — Men are good, women are neutral
Indifference curves are vertical lines. Freddy wants as many men as possible but is indifferent about the
number of women.
Chapter 4: Utility
Exercise 4.1
Solution: See table above.
Exercise 4.2
Solution:
Perfect substitutes with rate 2:3 require: U(2, 0) = U(0, 3)
a) U = 3x₁ + 2x₂ + 1000: U(2, 0) = 6 + 1000 = 1006; U(0, 3) = 6 + 1000 = 1006 ✓
This is a monotonic transformation of a perfect substitutes function.
b) U = (3x₁ + 2x₂)²: This is a monotonic transformation of (a). ✓
6
,c) U = min{3x₁, 2x₂}: This represents perfect complements, not perfect substitutes. ✗
For perfect complements: U(2, 0) = min{6, 0} = 0; U(0, 3) = min{0, 6} = 0 ✓ but this is coincidental.
Actually, testing: U(2, 0) = min{6, 0} = 0; U(0, 3) = min{0, 6} = 0; U(1, 1.5) = min{3, 3} = 3
This describes perfect complements, not perfect substitutes.
d) U = 30x₁ + 20x₂ - 10,000: U(2, 0) = 60 - 10,000 = -9,940; U(0, 3) = 60 - 10,000 = -9,940 ✓
This is a monotonic transformation of (a).
Answer: (c) does not describe Jonas’s preferences. It represents perfect complements, not perfect
substitutes.
Exercise 4.3
Solution:
Input: U(x₁, x₂) = 4x₁^(1/2) + x₂; initial bundle (81, 14)
Find the indifference curve through (81, 14):
𝑈(81,14) = 4 ⋅ √81 + 14 = 4 ⋅ 9 + 14 = 36 + 14 = 50
New bundle: x₁ = 81 + 40 = 121
For indifference: U(121, x₂) = 50
4 ⋅ √121 + 𝑥! = 50
4 ⋅ 11 + 𝑥! = 50
44 + 𝑥! = 50
𝑥! = 6
Change in cherries: Δx₂ = 14 - 6 = 8
Answer: Tom is willing to forgo 8 cherries for 40 extra nuts.
Remark: This exercise cannot be solved using MRS alone because the change (40 units) is too large. MRS
gives only the marginal rate of substitution at a specific bundle and applies only to infinitesimal changes.
Exercise 4.4
Solution A:
Input: Jan: 1 glass 0.5L (x₂) = 2 glasses 0.25L (x₁); Alex: 1 glass 0.5L ≈ 1 glass 0.25L
7
,For Jan (perfect substitutes with rate 2:1): - 1 unit of x₂ = 2 units of x₁ - U(x₁, x₂) = x₁ + 2x₂ - Check: U(2, 0) =
2; U(0, 1) = 2 ✓
For Alex (perfect substitutes with rate 1:1): - 1 unit of x₂ ≤ 1 unit of x₁ (he dislikes warm beer) - U(x₁, x₂) = x₁
+ x₂ - Check: U(1, 0) = 1; U(0, 1) = 1 ✓
Solution B:
Jan’s utility function represents: x₁ + 2x₂ (or any monotonic transformation)
• U(x₁, x₂) = 100x₁ + 200x₂: Check U(2, 0) = 200; U(0, 1) = 200 ✓ Yes
• U(x₁, x₂) = (5x₁ + 10x₂)²: Check U(2, 0) = (10)² = 100; U(0, 1) = (10)² = 100 ✓ Yes (monotonic
transformation)
• U(x₁, x₂) = x₁ + 3x₂: Check U(2, 0) = 2; U(0, 1) = 3 ✗ No (2 ≠ 3)
Answer: Yes, the first two functions describe Jan’s preferences. The third does not.
Exercise 4.5
Solution A:
Input: U(x_A, x_B) = x_A · x_B; bundle (40, 5)
U(40, 5) = 40 · 5 = 200
The indifference curve through (40, 5) gives all bundles where x_A · x_B = 200
Equation:
200
𝑥% =
𝑥&
Solution B:
The indifference curve is a rectangular hyperbola passing through (40, 5), (50, 4), (100, 2), etc.
Solution C:
Test the proposed exchange: (40, 5) → (15, 20)
U(15, 20) = 15 · 20 = 300 > 200
Answer: Yes, this exchange would make Charlie happier.
Maximum apples for 15 bananas while maintaining same utility:
U(x_A, 20) = 200
𝑥& ⋅ 20 = 200
8
, 𝑥& = 10
So the bundle (10, 20) gives the same utility as (40, 5).
Answer: For 15 extra bananas (from 5 to 20), Charlie is willing to give up a maximum of 40 - 10 = 30 apples.
Summary
Number of
Chapter Exercises Key Topics
2: Budget 4 Income effects, price effects, non-linear budgets, subsidies
Constraint
3: Preferences 4 Indifference curves, MRS, preference types (substitutes,
complements, bads, neutrals)
4: Utility 5 Utility functions, MU, MRS calculations, willingness to trade, Cobb-
Douglas
Total 13 Foundations of consumer choice theory
Chapter 2 — Consumer Choice & Individual Demand (H5, H6)
Solutions — WPO 2
Chapter 5 — Consumer Choice
Exercise 5.1
Solution via MRS-Condition:
First, find the marginal utilities: - MU₁ = ∂U/∂x₁ = 2x₁^(-1/2) - MU₂ = ∂U/∂x₂ = 1
Apply the MRS condition: -p₁/p₂ = -MU₁/MU₂
Substituting: 1/4 = 2x₁^(-1/2)
Solving: x₁^(-1/2) = 1/8 x₁^(1/2) = 8 x₁ = 64
From the budget constraint: 72 = 1(64) + 4x₂ x₂ = 2
Note: This is a quasilinear preference, where x₁ is independent of the budget.
Solution via Lagrange:
9