TARGETED WORKSHOP

Skill Repair & Drill

Six focused modules targeting the exact mechanics that break down under exam pressure.

① The Partial Derivative Problem

This is the single most important module. Every Lagrangian-based problem in this course requires you to take partial derivatives correctly. One wrong rule at this step breaks the entire solution. The rule is simple: when you differentiate with respect to $x$, everything that doesn't contain $x$ is a constant and vanishes.

The law to tattoo on your brain:
If $f(x,y) = x + 4\ln y$, then: $$\frac{\partial f}{\partial x} = 1 \qquad \frac{\partial f}{\partial y} = \frac{4}{y}$$ The term $4\ln y$ does NOT appear in $\partial f/\partial x$ — it has no $x$ in it. The term $x$ does NOT appear in $\partial f/\partial y$ — it has no $y$ in it.
⚠️ The exact error to avoid:
Writing $\frac{\partial}{\partial x}[x + 4\ln y] = 1 + 4\ln(y)$ — WRONG.
Writing $\frac{\partial}{\partial y}[x + 4\ln y] = x + \frac{4}{y}$ — WRONG.
Both errors come from carrying terms that belong to the OTHER variable.

Practice each one. Read the question, predict the answer, then reveal. Do this until it feels automatic.

$f(x,y) = 3x + 2\ln y$
Find $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$.
Show answer
$\dfrac{\partial f}{\partial x} = 3$ (the $2\ln y$ vanishes — no $x$)
$\dfrac{\partial f}{\partial y} = \dfrac{2}{y}$ (the $3x$ vanishes — no $y$)
Lagrangian: $\mathcal{L} = x + 4\ln y + \lambda(12 - x - py)$
Find $\dfrac{\partial \mathcal{L}}{\partial x}$.
Show answer
$\dfrac{\partial \mathcal{L}}{\partial x} = 1 + 0 + \lambda(0 - 1) = 1 - \lambda$
The $4\ln y$ term has zero $x$-content. The constraint gives $-\lambda \cdot 1$.
Same Lagrangian: $\mathcal{L} = x + 4\ln y + \lambda(12 - x - py)$
Find $\dfrac{\partial \mathcal{L}}{\partial y}$.
Show answer
$\dfrac{\partial \mathcal{L}}{\partial y} = 0 + \dfrac{4}{y} + \lambda(0 - p) = \dfrac{4}{y} - \lambda p$
The $x$ term is a constant w.r.t. $y$. The constraint gives $-\lambda p$.
$u(x,y) = x^{1/2}y^{1/2}$
Find $\dfrac{\partial u}{\partial x}$ and $\dfrac{\partial u}{\partial y}$.
Show answer
$\dfrac{\partial u}{\partial x} = \dfrac{1}{2}x^{-1/2}y^{1/2} = \dfrac{y^{1/2}}{2x^{1/2}}$
$\dfrac{\partial u}{\partial y} = \dfrac{1}{2}x^{1/2}y^{-1/2} = \dfrac{x^{1/2}}{2y^{1/2}}$
MRS $= \dfrac{\partial u/\partial x}{\partial u/\partial y} = \dfrac{y}{x}$
$\mathcal{L} = x^{1/2}y^{1/2} + \lambda(m - p_x x - p_y y)$
Find both $\dfrac{\partial \mathcal{L}}{\partial x}$ and $\dfrac{\partial \mathcal{L}}{\partial y}$, set to zero.
Show answer
$\dfrac{\partial \mathcal{L}}{\partial x} = \dfrac{y^{1/2}}{2x^{1/2}} - \lambda p_x = 0$
$\dfrac{\partial \mathcal{L}}{\partial y} = \dfrac{x^{1/2}}{2y^{1/2}} - \lambda p_y = 0$
Divide: $\dfrac{y}{x} = \dfrac{p_x}{p_y}$ → MRS = price ratio ✓
$u(x,y) = \ln x + \beta\ln y$
Write the Lagrangian and both FOCs for the UMP with prices $p_x, p_y$ and income $m$.
Show answer
$\mathcal{L} = \ln x + \beta\ln y + \lambda(m - p_x x - p_y y)$
$\partial/\partial x$: $\dfrac{1}{x} - \lambda p_x = 0 \Rightarrow \lambda = \dfrac{1}{p_x x}$
$\partial/\partial y$: $\dfrac{\beta}{y} - \lambda p_y = 0 \Rightarrow \lambda = \dfrac{\beta}{p_y y}$
Equate: $\dfrac{1}{p_x x} = \dfrac{\beta}{p_y y} \Rightarrow y = \dfrac{\beta p_x}{p_y}x$
Production: $y = L^{1/4}K^{1/4}$
Find $MP_L$ and $MP_K$.
Show answer
$MP_L = \dfrac{\partial y}{\partial L} = \dfrac{1}{4}L^{-3/4}K^{1/4}$
$MP_K = \dfrac{\partial y}{\partial K} = \dfrac{1}{4}L^{1/4}K^{-3/4}$
MRTS $= \dfrac{MP_L}{MP_K} = \dfrac{K}{L}$
$f(x,y) = x^2 + 3xy - 2y^2$
Find both partials. (Careful: $3xy$ contains both!)
Show answer
$\dfrac{\partial f}{\partial x} = 2x + 3y$ (the $3xy$ term differentiates to $3y$)
$\dfrac{\partial f}{\partial y} = 3x - 4y$ (the $3xy$ term differentiates to $3x$)
Cross-terms: both partials keep the OTHER variable as a coefficient.

② Quasilinear Utility — Full Template

Problem 1 Pattern
QUASILINEAR UMP

Brain hook: You love sushi ($y$) but you'd happily trade any sushi for cash ($x$, the numeraire). Your marginal utility from $x$ is just 1 — constant. That's what quasilinear means.

The Problem. A consumer has utility $u(x,y) = x + 4\ln y$. Prices: $p_x = 1$, $p_y = p$. Income $m = 12$.

(a) Derive Marshallian demands $x^*(p)$ and $y^*(p)$. Does demand for $y$ depend on income?
(b) Price of $y$ rises from $p=2$ to $p=4$. Find initial and final consumption bundles.
(c) Compute the compensating variation.
(d) Compare CV to the cost increase of keeping initial $y$ after the price rise.
Full Solution — parts (a) through (d)
1
Write the Lagrangian (correctly)
We maximize $u$ subject to the budget — the Lagrangian packages both.
$$\mathcal{L} = x + 4\ln y + \lambda(12 - x - py)$$
2
Take the three FOCs — carefully
Each FOC is a partial derivative of $\mathcal{L}$. Each variable $x$, $y$, $\lambda$ gets its own equation.
$$\frac{\partial \mathcal{L}}{\partial x} = \underbrace{1}_{\partial x/\partial x} + \underbrace{0}_{4\ln y\text{ has no }x} + \lambda \cdot \underbrace{(-1)}_{\partial(-x)/\partial x} = 1 - \lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = \underbrace{0}_{x\text{ has no }y} + \underbrace{\frac{4}{y}}_{\partial(4\ln y)/\partial y} + \lambda \cdot \underbrace{(-p)}_{\partial(-py)/\partial y} = \frac{4}{y} - \lambda p = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 12 - x - py = 0$$
⚠️ The critical trap: $4\ln y$ does NOT contribute to $\partial\mathcal{L}/\partial x$. The $x$ term does NOT contribute to $\partial\mathcal{L}/\partial y$. These terms are constants with respect to the other variable.
3
Read off $\lambda$ and solve for $y^*$
The first FOC gives $\lambda$ directly. The second FOC then solves for $y^*$.
From FOC(x): $\lambda = 1$

Plug into FOC(y): $\dfrac{4}{y} = \lambda p = p$ $$\boxed{y^*(p) = \frac{4}{p}}$$ Key observation: No $m$ appears. The demand for $y$ is income-independent. This is the signature of quasilinear utility.
4
Use the budget to find $x^*$
We have $y^*$; substitute into the binding budget constraint to get $x^*$.
$x + py = 12 \Rightarrow x^* = 12 - p \cdot y^* = 12 - p \cdot \frac{4}{p} = 12 - 4$ $$\boxed{x^*(p) = 8}$$ Also income-independent! (Because the "fixed spend" on $y$ is always $p \cdot (4/p) = 4$, regardless of $p$.)
💡 The quasilinear pattern: When $u = x + v(y)$ and $p_x = 1$, the FOC for $y$ pins down $y^*$ from $v'(y^*) = p_y$. This is always independent of income. Then $x^*$ soaks up whatever income is left. Think of $x$ as "savings" and $y$ as a fixed-quantity purchase.
5
Part (b): Plug in $p = 2$ then $p = 4$
No new technique needed — just evaluate the demand functions.
Initial ($p = 2$): $y_0^* = 4/2 = 2$, $x_0^* = 8$ → bundle $(8,\ 2)$
Final ($p = 4$): $y_1^* = 4/4 = 1$, $x_1^* = 8$ → bundle $(8,\ 1)$

Observe: $x$ doesn't move at all. All adjustment happens in $y$ (cut in half).
6
Part (c): Compensating Variation via the demand integral
CV = area under the (Hicksian = Marshallian) demand curve between the two prices. Quasilinear utility is special: Hicksian and Marshallian demands coincide for $y$ (no income effect), so we can just integrate the Marshallian demand.
$$CV = \int_{p_0}^{p_1} y^*(p)\, dp = \int_2^4 \frac{4}{p}\, dp = \Big[4\ln p\Big]_2^4$$ $$= 4\ln 4 - 4\ln 2 = 4\ln\!\left(\frac{4}{2}\right) = 4\ln 2 \approx 2.77$$
⚠️ Most common wrong approach: Trying to integrate the MRS or an unsimplified expression. Once you have $y^*(p) = 4/p$, the integral is immediate — it's just $\int 4/p\, dp = 4\ln p$.
7
Part (d): Compare to forced-quantity cost increase
If the consumer were frozen at $y_0 = 2$ and forced to pay the new price, how much extra would they spend? Compare that to CV.
Cost of $y_0 = 2$ at new price: $p_1 \cdot y_0 = 4 \times 2 = 8$
Cost of $y_0 = 2$ at old price: $p_0 \cdot y_0 = 2 \times 2 = 4$
Cost increase: $\Delta = 8 - 4 = 4$

$CV = 4\ln 2 \approx 2.77 < 4 = \Delta$

CV is smaller. The consumer doesn't need full compensation because she can substitute — she'll buy less $y$ at the higher price and adjust $x$ upward. The option to adjust is worth something; it reduces the required compensation.
💡 The intuition: CV < forced-cost-increase whenever the consumer can substitute. If she were frozen at $y_0$, she'd need $\Delta = 4$ extra dollars. But since she can and will cut $y$ to 1, she only needs $4\ln 2 \approx 2.77$ extra to reach the same utility.

🎯 Try it yourself — same structure, different function

$u(x,y) = x + 6\ln y$, $p_x = 1$, $p_y = p$, $m = 20$.
Find $x^*(p)$, $y^*(p)$, compute bundles at $p=3$ and $p=6$, and compute CV.

Show answer
FOC(x): $\lambda = 1$. FOC(y): $6/y = p \Rightarrow y^* = 6/p$.
Budget: $x^* = 20 - p(6/p) = 20 - 6 = 14$.

At $p = 3$: $(14,\ 2)$. At $p = 6$: $(14,\ 1)$.

$CV = \int_3^6 \frac{6}{p}\,dp = [6\ln p]_3^6 = 6\ln 6 - 6\ln 3 = 6\ln 2 \approx 4.16$

③ Producer Theory — The Complete Chain

Problem 2 Pattern
CMP → C(y) → MC → SUPPLY → EQUILIBRIUM

Brain hook: You run a factory. First minimize your costs for any given output level. That gives $C(y)$. From $C(y)$ derive MC and AC. MC = P tells you supply. Then clear the market.

The Problem. A firm has $y = L^{1/4}K^{1/4}$, input prices $w = r = 1$, fixed cost $F = 2$. Market demand $Q^D = 36 - p$.

(a) Derive cost function $C(y)$.
(b) Derive MC and AC. Find the supply function. Which prices induce production?
(c) $N = 4$ firms. Find competitive equilibrium $(p^*, Q^*, y^*, \pi^*)$.
(d) Free entry. Find equilibrium $(p^*, Q^*, y^*, N^*)$.
Full Solution — all four parts
1
Cost Minimization — find optimal input ratio
At the cost minimum, MRTS = input price ratio. This pins down how L and K relate.
$MP_L = \tfrac{1}{4}L^{-3/4}K^{1/4}$, $MP_K = \tfrac{1}{4}L^{1/4}K^{-3/4}$

$$\text{MRTS} = \frac{MP_L}{MP_K} = \frac{K}{L} = \frac{w}{r} = \frac{1}{1} = 1 \implies K^* = L^*$$
2
Substitute into production constraint to get $L^*(y)$
With $K = L$, the production function simplifies. Solve for how much $L$ you need to produce $y$.
$y = L^{1/4}K^{1/4} = L^{1/4}L^{1/4} = L^{1/2}$

$\implies L^* = y^2$, $K^* = y^2$
3
Write the cost function
Total cost = variable inputs + fixed cost. Plug in optimal $L^*, K^*$.
$$C(y) = w \cdot L^* + r \cdot K^* + F = 1 \cdot y^2 + 1 \cdot y^2 + 2$$ $$\boxed{C(y) = 2y^2 + 2}$$
4
Derive MC and AC, find minimum AC
The firm produces only if price covers average cost. Minimum AC gives the shut-down price threshold.
$MC(y) = C'(y) = 4y$
$AC(y) = C(y)/y = 2y + 2/y$

Min AC: $\dfrac{d(AC)}{dy} = 2 - \dfrac{2}{y^2} = 0 \implies y = 1$
$AC_{\min} = 2(1) + 2/1 = 4$

Shut-down rule: Produce if and only if $p \geq AC_{\min} = 4$.
5
Individual supply function
Competitive firm sets $MC = p$. Solve for $y$ in terms of $p$.
$4y = p \implies y^s(p) = \dfrac{p}{4}$ for $p \geq 4$, else $y = 0$.
6
Part (c): N = 4 firms, market equilibrium
Market supply = N × individual supply. Set equal to demand.
$Q^S = 4 \cdot \dfrac{p}{4} = p$

Market clearing: $Q^S = Q^D \implies p = 36 - p \implies 2p = 36$ $$p^* = 18, \quad Q^* = 18, \quad y^* = 18/4 = 4.5$$ Profit per firm: $\pi^* = p^*y^* - C(y^*) = 18(4.5) - 2(4.5)^2 - 2 = 81 - 40.5 - 2 = 38.5 > 0$
💡 Positive profit → entry will happen. At $p^* = 18 > AC_{\min} = 4$, each firm earns $38.5$. That's the signal for new firms to enter.
7
Part (d): Free entry — zero-profit equilibrium
Entry continues until profit = 0. Two conditions pin down $p^*$ and $y^*$: (1) $\pi = 0$ and (2) $MC = p$.
Condition 1 — MC = P: $p = 4y$
Condition 2 — $\pi = 0$: $py - 2y^2 - 2 = 0$

Substitute (1) into (2): $4y \cdot y - 2y^2 - 2 = 0 \implies 4y^2 - 2y^2 = 2 \implies 2y^2 = 2$ $$y^* = 1, \quad p^* = 4y^* = 4$$ $Q^* = 36 - 4 = 32$, $N^* = Q^*/y^* = 32$ firms
💡 The zero-profit condition means P = AC$_{\min}$. Firms produce at exactly the efficient scale where $AC = MC$. Price equals the bottom of the U-shaped AC curve. This is the long-run competitive equilibrium.
⚠️ Common error on zero-profit condition: Writing "zero profit condition: $MC = P$." That's actually the profit-maximization condition, not zero profit. Zero profit means $\pi = py - C(y) = 0$. Both conditions together are needed.

🎯 Try it — same structure, different production function

Firm has $y = L^{1/2}K^{1/2}$, $w = 4$, $r = 1$, $F = 9$. Demand $Q^D = 40 - p$.
Find $C(y)$, $MC$, $AC_{\min}$. With free entry, find $p^*, y^*, N^*$.

Show answer
MRTS: $K/L = w/r = 4 \Rightarrow K = 4L$. Production: $y = L^{1/2}(4L)^{1/2} = 2L$, so $L^* = y/2$, $K^* = 2y$.
$C(y) = 4(y/2) + 1(2y) + 9 = 2y + 2y + 9 = 4y + 9$. (Linear VC + fixed cost.)
$MC = 4$. $AC = 4 + 9/y$.
Min AC: $\frac{d}{dy}(4 + 9/y) = -9/y^2 \neq 0$ for finite $y$ — AC is always decreasing (no interior min). Entry drives $N \to \infty$? Actually for linear MC with fixed cost, $p^* = MC = 4$ at free entry (since min AC → 4 as $y \to \infty$ — but this is increasing returns to scale, so the competitive equilibrium with free entry needs careful treatment). This is an important case to flag.

④ Walrasian GE — Linear + Cobb-Douglas

Problem 3 Pattern
NON-STANDARD DEMAND · CORNER SOLUTIONS · MARKET CLEARING

Brain hook: One consumer treats $x$ and $y$ as perfect substitutes (just counts total goods). The other has standard Cobb-Douglas preferences. When one consumer is linear, their demand is all-or-nothing depending on the price ratio — this is the key twist.

The Problem. $u_A = x_A + y_A$, $\omega_A = (4,2)$. $u_B = x_By_B$, $\omega_B = (2,4)$. Normalize $p_y = 1$, let $p_x = p$.

(a) Show consumers want to trade away from endowment $E$. Use MRS.
(b) Derive each consumer's Walrasian demand as a function of $p$.
(c) Find $p^*$ and equilibrium allocations.
(d) Characterize the Pareto set analytically.
Full Solution
1
Part (a): Check MRS at endowment
If MRS ≠ price ratio, consumers are not at their optimum — they want to trade.
MRS$^A = u_{Ax}/u_{Ay} = 1/1 = 1$ (constant — linear utility always has MRS = 1)

MRS$^B = u_{Bx}/u_{By} = y_B/x_B$. At endowment, B has $(2, 4)$:
$\text{MRS}^B = 4/2 = 2$

Since MRS$^B = 2 > 1 = $ MRS$^A$: B values $x$ more than A does. B will give up $y$ for $x$; A will give up $x$ for $y$. Both gain from trading at any price $p \in (1, 2)$.
2
Part (b): Walrasian demand for A — the linear case
With linear utility $u_A = x + y$, A always buys only the cheaper good (corner solution). The price ratio $p$ determines which corner.
Income: $m_A = p \cdot 4 + 1 \cdot 2 = 4p + 2$

A's problem: maximize $x_A + y_A$ s.t. $px_A + y_A = 4p + 2$.

A's MRS = 1. The good with the lower price per util gets all the spending: $$x_A^*(p) = \begin{cases} \dfrac{4p+2}{p} = 4 + \dfrac{2}{p} & \text{if } p < 1 \text{ (x cheaper)} \\ \text{any } x_A \in \left[0,\, \frac{4p+2}{p}\right] & \text{if } p = 1 \text{ (indifferent)} \\ 0 & \text{if } p > 1 \text{ (y cheaper)} \end{cases}$$ $$y_A^*(p) = \begin{cases} 0 & \text{if } p < 1 \\ \text{any } y_A \in \left[0,\, 4p+2\right] & \text{if } p = 1 \\ 4p+2 & \text{if } p > 1 \end{cases}$$
💡 Linear utility = bang-for-buck shopping. If $p_x < p_y$ (i.e., $p < 1$), A spends everything on the cheaper $x$. If $p_x > p_y$, spend everything on $y$. If equal, A is indifferent — here the exact split is determined by market clearing.
3
Part (b): Walrasian demand for B — Cobb-Douglas
Standard equal-share Cobb-Douglas: half income on each good.
Income: $m_B = p \cdot 2 + 1 \cdot 4 = 2p + 4$
$$x_B^*(p) = \frac{m_B}{2p} = \frac{2p+4}{2p} = 1 + \frac{2}{p}$$ $$y_B^*(p) = \frac{m_B}{2} = p + 2$$
4
Part (c): Market clearing — find $p^*$
Try the case $p = 1$ first (the natural candidate when MRS^A = 1). Verify it clears both markets.
At $p = 1$: A is indifferent, so let A choose $x_A^*$ freely (to be determined by clearing).
$x_B^*(1) = 1 + 2 = 3$, $y_B^*(1) = 1 + 2 = 3$

Market clearing for $x$: $x_A^* + 3 = 6 \Rightarrow x_A^* = 3$
A's budget at $p=1$: $x_A^* + y_A^* = m_A = 4(1)+2 = 6$, so $y_A^* = 6 - 3 = 3$

Check $y$: $y_A^* + y_B^* = 3 + 3 = 6 = \bar{y}$ ✓ $$\boxed{p^* = 1, \quad (x_A^*, y_A^*) = (3,3), \quad (x_B^*, y_B^*) = (3,3)}$$
5
Part (d): Pareto set characterization
Pareto efficiency: no reallocation makes everyone better off. For interior allocations, this requires equal MRS across consumers.
Interior Pareto condition: $\text{MRS}^A = \text{MRS}^B$
$1 = y_B/x_B \Rightarrow y_B = x_B$

In terms of A's allocation ($x_B = 6 - x_A$, $y_B = 6 - y_A$): $$6 - y_A = 6 - x_A \implies \boxed{y_A = x_A, \quad x_A \in [0,6]}$$ The contract curve is the 45° diagonal of the Edgeworth box (the line from $O_A$ to $O_B$ where $y_A = x_A$).
💡 Linear utility makes the Pareto set a specific line, not a band. Because MRS$^A$ is always exactly 1 (never changes with quantities), the contract curve is pinned wherever MRS$^B = 1$, which is exactly the diagonal $x_B = y_B$.

🎯 Try it — Pareto improvements over the endowment

Using the same economy ($u_A = x_A + y_A$, $\omega_A = (4,2)$, $u_B = x_By_B$, $\omega_B = (2,4)$): which allocations on the contract curve ($y_A = x_A$) are Pareto improvements over $E$? Solve explicitly.

Show answer
At endowment: $u_A(E) = 4+2 = 6$, $u_B(E) = 2 \times 4 = 8$.
On contract curve: $x_A \in [0,6]$, $y_A = x_A$, so $u_A = 2x_A$.
$x_B = 6 - x_A$, $y_B = 6 - x_A$, so $u_B = (6-x_A)^2$.

Pareto improvement: $2x_A \geq 6$ AND $(6-x_A)^2 \geq 8$ with at least one strict.
$2x_A \geq 6 \Rightarrow x_A \geq 3$.
$(6-x_A)^2 \geq 8 \Rightarrow 6 - x_A \geq 2\sqrt{2} \Rightarrow x_A \leq 6 - 2\sqrt{2} \approx 3.17$.

Pareto improvements on the contract curve: $x_A \in [3,\ 6 - 2\sqrt{2}]$, $y_A = x_A$. This is a small segment of the diagonal — only a narrow band of trades are mutually beneficial starting from this endowment.

⑤ Compensating Variation — Three Methods

Core Welfare Concept
What CV means: The price of $y$ rises. How much extra money do we need to give the consumer so she's exactly as happy as before? That's the CV.
All three methods — when to use each
M1
Method 1: Expenditure function difference
The cleanest method when you have the expenditure function $e(p_x, p_y, u)$.
$$CV = e(p_x, p_y^1, u_0) - e(p_x, p_y^0, u_0)$$ Cost of achieving old utility $u_0$ at new prices minus cost at old prices.

For the exam problem ($u = x + 4\ln y$, $p_x = 1$):
Expenditure function (from Hicksian: $y^h = 4/p_y$): $e(1, p_y, u_0) = u_0 + 4 - 4\ln(4/p_y) = u_0 - 4\ln 4 + 4\ln p_y + 4$

$CV = e(1, 4, u_0) - e(1, 2, u_0) = [4\ln 4 + 4] - [4\ln 2 + 4] = 4\ln(4/2) = 4\ln 2$
M2
Method 2: Area under Hicksian demand
CV = area between two price levels under the Hicksian (compensated) demand curve for the good whose price changed.
$$CV = \int_{p_y^0}^{p_y^1} h_y(p_x, p_y, u_0)\, dp_y$$ For quasilinear utility: $h_y = y^* = 4/p_y$ (Hicksian = Marshallian — no income effect).
$$CV = \int_2^4 \frac{4}{p}\, dp = 4\ln 4 - 4\ln 2 = 4\ln 2$$
M3
Method 3: Area under Marshallian demand (approximation, exact for quasilinear)
For quasilinear utility, Hicksian = Marshallian, so area under Marshallian demand is exact. For other utilities, it's only an approximation.
Same integral, same answer. Key caveat: for Cobb-Douglas or other non-quasilinear utilities, this gives an approximation — you'd need the Hicksian demand explicitly (via Shephard's Lemma on the expenditure function) for the exact answer.
💡 When Hicksian = Marshallian: Only for quasilinear utility (zero income effect). In all other cases, use the expenditure function method for exactness.

⑥ First and Second Welfare Theorems — True/False Drill

Theory Foundation
FWT: Every Walrasian equilibrium is Pareto efficient.
SWT: Every Pareto-efficient allocation can be supported as a Walrasian equilibrium — but possibly after redistributing endowments (lump-sum transfers).
T or F: "Every Pareto-efficient allocation can be sustained as a Walrasian equilibrium from the original endowment."
Show answer
FALSE. This confuses FWT with SWT. The SWT says every Pareto allocation can be a Walrasian equilibrium, but it may require redistributing the endowment first. From a fixed endowment, there is typically only one Walrasian equilibrium — not all Pareto-efficient points.
T or F: "The Walrasian equilibrium in the Edgeworth box is always on the contract curve."
Show answer
TRUE. By the FWT, every Walrasian equilibrium is Pareto efficient. The contract curve is the set of Pareto-efficient interior allocations. So the Walrasian equilibrium allocation is on the contract curve (assuming it's interior).
T or F: "If MRS$^A \neq $ MRS$^B$ at an allocation, that allocation is Pareto inefficient."
Show answer
TRUE (for interior allocations). If MRS$^A \neq$ MRS$^B$, there's a mutually beneficial trade — a reallocation that makes at least one person better off without hurting the other. Hence the allocation is not Pareto efficient. (At boundary/corner allocations, we need to check more carefully.)
T or F: "The Pareto set and the contract curve are the same thing."
Show answer
APPROXIMATELY TRUE but imprecise. The contract curve is typically the interior portion of the Pareto set (where MRS$^A$ = MRS$^B$ at interior points). The full Pareto set may also include boundary allocations. In practice at the 101A level, contract curve ≈ Pareto set of interior allocations.