2026
INTERMEDIATE MICROECONOMICS
SUMMARY EXERCISE SOLUTIONS
KOEN HANEGREEFS
VUB
,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.
Ex. 2.2 — Amy’s budget line
1
,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).
Ex. 2.3a — Budget set with quantity discount
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.
Ex. 2.3b — Budget set with €6,000 connection fee (parallel shift)
2
,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.
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.
3
, Ex. 2.4 — Four subsidy scenarios
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
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.
4
, Ex. 3.1c — Lunch time vs other goods
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).
5
,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
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.
6
, Ex. 3.4 — Five preference types
Chapter 4: Utility
Exercise 4.1
Solution:
Marginal utility is the partial derivative: 𝑀𝑈% = 𝜕𝑈/𝜕𝑥% . The MRS is −𝑀𝑈" /𝑀𝑈! .
• U = 2x₁ + 3x₂ → MU₁ = 2, MU₂ = 3, MRS = −2/3
• U = 4x₁ + 6x₂ → MU₁ = 4, MU₂ = 6, MRS = −2/3 (same preferences as line above; positive linear
transformation)
• U = ax₁ + bx₂ → MU₁ = a, MU₂ = b, MRS = −a/b
• U = 2x₁^(1/2) + x₂ → MU₁ = x₁^(−1/2), MU₂ = 1, MRS = −x₁^(−1/2) (quasilinear in x₂)
• U = ln x₁ + x₂ → MU₁ = 1/x₁, MU₂ = 1, MRS = −1/x₁ (quasilinear in x₂)
• U = v(x₁) + x₂ → MU₁ = v′(x₁), MU₂ = 1, MRS = −v′(x₁) (general quasilinear form)
• U = x₁ · x₂ → MU₁ = x₂, MU₂ = x₁, MRS = −x₂/x₁ (Cobb-Douglas, exponents 1,1)
• U = x₁^a · x₂^b → MU₁ = a·x₁(a−1)·x₂b, MU₂ = b·x₁a·x₂(b−1), MRS = −(a·x₂)/(b·x₁)
• U = (x₁ + 2)(x₂ + 1) → MU₁ = x₂ + 1, MU₂ = x₁ + 2, MRS = −(x₂ + 1)/(x₁ + 2) (shifted Cobb-Douglas)
• U = (x₁ + a)(x₂ + b) → MU₁ = x₂ + b, MU₂ = x₁ + a, MRS = −(x₂ + b)/(x₁ + a) (general shifted form)
• U = x₁^a + x₂^a → MU₁ = a·x₁^(a−1), MU₂ = a·x₂^(a−1), MRS = −(x₁/x₂)^(a−1) (CES-like)
Key takeaways: - Linear utility (perfect substitutes) → MRS is constant. - Quasilinear (linear in one good) →
MRS depends only on the other good. - Cobb-Douglas → MRS depends on the ratio x₂/x₁.
7
,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). ✓
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.
8
, 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
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.
9
INTERMEDIATE MICROECONOMICS
SUMMARY EXERCISE SOLUTIONS
KOEN HANEGREEFS
VUB
,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.
Ex. 2.2 — Amy’s budget line
1
,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).
Ex. 2.3a — Budget set with quantity discount
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.
Ex. 2.3b — Budget set with €6,000 connection fee (parallel shift)
2
,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.
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.
3
, Ex. 2.4 — Four subsidy scenarios
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
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.
4
, Ex. 3.1c — Lunch time vs other goods
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).
5
,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
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.
6
, Ex. 3.4 — Five preference types
Chapter 4: Utility
Exercise 4.1
Solution:
Marginal utility is the partial derivative: 𝑀𝑈% = 𝜕𝑈/𝜕𝑥% . The MRS is −𝑀𝑈" /𝑀𝑈! .
• U = 2x₁ + 3x₂ → MU₁ = 2, MU₂ = 3, MRS = −2/3
• U = 4x₁ + 6x₂ → MU₁ = 4, MU₂ = 6, MRS = −2/3 (same preferences as line above; positive linear
transformation)
• U = ax₁ + bx₂ → MU₁ = a, MU₂ = b, MRS = −a/b
• U = 2x₁^(1/2) + x₂ → MU₁ = x₁^(−1/2), MU₂ = 1, MRS = −x₁^(−1/2) (quasilinear in x₂)
• U = ln x₁ + x₂ → MU₁ = 1/x₁, MU₂ = 1, MRS = −1/x₁ (quasilinear in x₂)
• U = v(x₁) + x₂ → MU₁ = v′(x₁), MU₂ = 1, MRS = −v′(x₁) (general quasilinear form)
• U = x₁ · x₂ → MU₁ = x₂, MU₂ = x₁, MRS = −x₂/x₁ (Cobb-Douglas, exponents 1,1)
• U = x₁^a · x₂^b → MU₁ = a·x₁(a−1)·x₂b, MU₂ = b·x₁a·x₂(b−1), MRS = −(a·x₂)/(b·x₁)
• U = (x₁ + 2)(x₂ + 1) → MU₁ = x₂ + 1, MU₂ = x₁ + 2, MRS = −(x₂ + 1)/(x₁ + 2) (shifted Cobb-Douglas)
• U = (x₁ + a)(x₂ + b) → MU₁ = x₂ + b, MU₂ = x₁ + a, MRS = −(x₂ + b)/(x₁ + a) (general shifted form)
• U = x₁^a + x₂^a → MU₁ = a·x₁^(a−1), MU₂ = a·x₂^(a−1), MRS = −(x₁/x₂)^(a−1) (CES-like)
Key takeaways: - Linear utility (perfect substitutes) → MRS is constant. - Quasilinear (linear in one good) →
MRS depends only on the other good. - Cobb-Douglas → MRS depends on the ratio x₂/x₁.
7
,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). ✓
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.
8
, 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
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.
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