Section 10

When One Agent Has Linear Preferences

Section 9 paired two Cobb-Douglas agents, both of whom always want a strictly interior bundle. This example swaps one agent for perfect substitutes — a linear utility — which behaves very differently: its demand is all-or-nothing in price, and it ends up pinning the equilibrium price by itself. We also give both agents a mix of the two goods to start, so no one begins with nothing.

The primitives

Agent $A$ has Cobb-Douglas preferences with equal weights; agent $B$ has linear preferences (perfect substitutes) that value good 1 twice as much as good 2 at the margin.

u^A(x_1, x_2) = x_1^{0.5}\, x_2^{0.5}, \qquad u^B(x_1, x_2) = 2\,x_1 + x_2

\alpha_A = 0.5

, and

B

's marginal rate of substitution is the constant

\mathrm{MRS}^B = 2

Both agents hold some of each good — these are interior endowments, not the "one agent owns everything" setup of §9.

\omega^A = (\omega^A_1, \omega^A_2) = (8,\ 4), \qquad \omega^B = (\omega^B_1, \omega^B_2) = (4,\ 8)

Tastes: $A$ : Cobb-Douglas, $\alpha_A = 0.5$ . $B$ : linear, $u^B = 2x_1 + x_2$ .
Total endowments: $\Omega_1 = \omega^A_1 + \omega^B_1 = 12$ , $\Omega_2 = \omega^A_2 + \omega^B_2 = 12$ — a square box.
Numéraire: Good 2, so $p_2 = 1$ and we solve only for $p_1$ .

Step 1 — Agent A's demand (Cobb-Douglas)

$A$ maximises $u^A = x_1^{0.5} x_2^{0.5}$ subject to the budget constraint $p_1 x_1 + x_2 = m^A$ , where wealth is the market value of the endowment, $m^A = p_1 \omega^A_1 + \omega^A_2 = 8\,p_1 + 4$ . Because $A$ 's preferences are smooth and strictly convex, the optimum is the interior point where the indifference curve is tangent to the budget line — that is, where the marginal rate of substitution equals the price ratio. We can write that condition down directly, without a Lagrangian:

Step 1
$\mathrm{MRS}^A = \frac{\partial u^A / \partial x_1}{\partial u^A / \partial x_2} = \frac{0.5\,x_1^{-0.5} x_2^{0.5}}{0.5\,x_1^{0.5} x_2^{-0.5}} = \frac{x_2}{x_1}$
The marginal rate of substitution is the ratio of marginal utilities. For equal-weighted Cobb-Douglas it simplifies to $x_2 / x_1$ .
Step 2
$\frac{x_2}{x_1} = p_1 \;\Longrightarrow\; x_2 = p_1 x_1$
Tangency: set $\mathrm{MRS}^A$ equal to the price ratio $p_1/p_2 = p_1$ . This is the optimality condition — buying or selling along the budget line cannot improve utility once it holds.
Step 3
$p_1 x_1 + p_1 x_1 = m^A \;\Longrightarrow\; 2 p_1 x_1 = m^A$
Substitute $x_2 = p_1 x_1$ into the budget constraint.
Step 4
$x^A_1(p_1) = \frac{m^A}{2 p_1} = \frac{8 p_1 + 4}{2 p_1}, \qquad x^A_2(p_1) = \frac{m^A}{2} = \frac{8 p_1 + 4}{2}$
Solve for each good. With $\alpha_A = 0.5$ the agent splits wealth evenly: half on good 1, half on good 2 — exactly the §5 formula $x^A_1 = \alpha_A m^A / p_1$ , $x^A_2 = (1-\alpha_A) m^A$ .

Step 2 — Agent B's demand (perfect substitutes)

The tangency trick from Step 1 fails here. $B$ 's indifference curves are straight lines with constant slope $\mathrm{MRS}^B = 2$ , so the condition $\mathrm{MRS}^B = p_1$ holds at no point when $p_1 \ne 2$ and at every point when $p_1 = 2$ — never at a single interior bundle. Instead we maximise directly. Solve the budget for $x_2$ and note that feasibility ( $x_1, x_2 \ge 0$ ) confines $x_1$ to a finite interval:

x_2 = m^B - p_1 x_1, \qquad 0 \le x_1 \le \frac{m^B}{p_1}

The endpoints are the corners of the budget line:

x_1 = 0

(all good 2) and

x_1 = m^B/p_1

(all good 1, since then

x_2 = 0

Step 1
$u^B = 2 x_1 + x_2 = 2 x_1 + (m^B - p_1 x_1) = m^B + (2 - p_1)\, x_1$
Substitute $x_2$ into utility. Along the budget line $u^B$ is a straight line in $x_1$ with slope $(2 - p_1)$ — so its maximum sits at one end of the interval $[0,\, m^B/p_1]$ , never in the middle.
Step 2
$p_1 < 2:\ \ \text{slope } (2 - p_1) > 0 \;\Longrightarrow\; \text{take } x_1 = \frac{m^B}{p_1} \;\Rightarrow\; \left(x^B_1, x^B_2\right) = \left(\tfrac{m^B}{p_1},\, 0\right)$
$u^B$ rises in $x_1$ , so pick the right endpoint — spend everything on good 1.
Step 3
$p_1 > 2:\ \ \text{slope } (2 - p_1) < 0 \;\Longrightarrow\; \text{take } x_1 = 0 \;\Rightarrow\; \left(x^B_1, x^B_2\right) = \left(0,\, m^B\right)$
$u^B$ falls in $x_1$ , so pick the left endpoint — buy only good 2.
Step 4
$p_1 = 2:\ \ \text{slope } (2 - p_1) = 0 \;\Longrightarrow\; u^B = m^B \ \text{ for all } x_1 \in \left[0,\, \tfrac{m^B}{p_1}\right]$
The slope vanishes, so utility is the same at every bundle on the budget line. $B$ is indifferent and becomes the residual trader.

This is the bang-bang rule: a perfect-substitutes agent buys only the good with the higher utility-per-dollar, comparing $b_1/p_1 = 2/p_1$ against $b_2/p_2 = 1$ — equivalently, comparing its constant $\mathrm{MRS}^B = 2$ against the price ratio $p_1$ .

Step 3 — The linear agent pins the price

The only price that can clear both markets is $p_1^* = 2$ . The two specialised cases each fail to clear, which forces the price to $B$ 's $\mathrm{MRS}$ :

Step 1
$p_1 < 2:\quad x^A_1 + x^B_1 = \left(4 + \tfrac{2}{p_1}\right) + \frac{4p_1 + 8}{p_1} > 12 = \Omega_1$
$B$ dumps all wealth into good 1, so combined demand for good 1 exceeds supply — excess demand pushes $p_1$ up.
Step 2
$p_1 > 2:\quad x^A_1 + x^B_1 = \left(4 + \tfrac{2}{p_1}\right) + 0 < 12 = \Omega_1$
$B$ abandons good 1, so $A$ 's demand alone falls short of supply — excess supply pushes $p_1$ down.
Step 3
$\;\Longrightarrow\; p_1^* = 2$
Squeezed from both sides, the price settles exactly at $B$ 's marginal rate of substitution. At $p_1^* = 2$ , $B$ is indifferent and absorbs whatever $A$ leaves.

Step 4 — The Walrasian allocation

Evaluate $A$ 's demand at $p_1^* = 2$ . Wealth is $m^A = 8 \cdot 2 + 4 = 20$ , so:

Step 1
$x^{A*}_1 = \frac{0.5 \cdot 20}{2} = 5, \qquad x^{A*}_2 = 0.5 \cdot 20 = 10$
$A$ sells 3 units of good 1 (8 → 5) and buys 6 units of good 2 (4 → 10).
Step 2
$x^{B*}_1 = \Omega_1 - x^{A*}_1 = 12 - 5 = 7, \qquad x^{B*}_2 = \Omega_2 - x^{A*}_2 = 12 - 10 = 2$
$B$ takes the residual: buys 3 units of good 1 (4 → 7), sells 6 units of good 2 (8 → 2).

x^{A*} = (5,\ 10), \qquad x^{B*} = (7,\ 2)

Walrasian allocation

Step 5 — Gains from trade

Here the two preference types part ways. $A$ , who wants balance, gains strictly from trade. $B$ , indifferent along the whole budget line, ends up exactly as well off as at the endowment:

Step 1
$u^A(\omega^A) = \sqrt{8 \cdot 4} = \sqrt{32} \approx 5.66 \quad\longrightarrow\quad u^A(x^{A*}) = \sqrt{5 \cdot 10} = \sqrt{50} \approx 7.07$
$A$ is strictly better off — the Cobb-Douglas agent captures the gains.
Step 2
$u^B(\omega^B) = 2 \cdot 4 + 8 = 16 \quad\longrightarrow\quad u^B(x^{B*}) = 2 \cdot 7 + 2 = 16$
$B$ is exactly indifferent: along the budget line $2x_1 + x_2 = 16$ , utility is constant at the wealth level.

This is the signature of a marginal linear trader: its utility equals its wealth, $u^B = m^B$ , and trade at the market price neither helps nor hurts it. Individual rationality still holds — weakly for $B$ , strictly for $A$ — so the outcome is a genuine equilibrium with no one made worse off.

Seeing both agents

The two diagrams below show each agent in their own coordinates. $A$ 's Cobb-Douglas indifference curve is tangent to the budget line at a single interior point — the usual smooth optimum. $B$ 's linear indifference curves are parallel straight lines with the same slope $-2$ as the budget line, so the budget line coincides with one of them: every point on it is optimal, which is why $B$ is content to sit at the residual bundle.

Agent A — Cobb-Douglas

Tangency at $x^{A*} = (5, 10)$

Budget line

2x_1 + x_2 = 20

through

\omega^A = (8,4)

; the indifference curve

u^A = \sqrt{50}

is tangent at

x^{A*}

Agent B — perfect substitutes

Budget line coincides with the $u^B = 16$ line

B's indifference curves are parallel lines of slope

-2

. The budget line

2x_1 + x_2 = 16

is itself the

u^B = 16

curve, so

\omega^B

and

x^{B*}

lie on the same indifference line.