Answer

Solution

1. 1

   Normalise $p_x = p_y = 1$, income $w > 0$. The quasi-linear FOC is

2. 2
   $$\frac{\partial U/\partial x}{\partial U/\partial y} = \frac{p_x}{p_y} \iff \frac{1}{(1+x)^2} = 1.$$
3. 3

   This gives $1 + x = 1$, i.e.\ $x = 0$. But the marginal utility of $x$ at $x = 0$ is $1$, equal to $MU_y$, so the consumer is indifferent at the corner. Check: if $x > 0$ tiny, $\partial U/\partial x = 1/(1+x)^2 < 1 = MU_y$, so spending on $x$ is sub-optimal at the margin.

4. 4

   However, consider the interior carefully. With $w > 0$ and $p_x = p_y$, allocate $\epsilon > 0$ to $x$ and $w - \epsilon$ to $y$. Utility is $\epsilon/(1+\epsilon) + (w-\epsilon)$. Differentiating w.r.t.\ $\epsilon$: $1/(1+\epsilon)^2 - 1 < 0$ for $\epsilon > 0$. So utility falls as $\epsilon$ rises from $0$. Optimum: $x = 0$, $y = w$.

5. 5

   \textit{Re-examination.} With $MU_x(0)/p_x = 1 = MU_y/p_y$, the FOC holds with equality at $x = 0$. For $x > 0$, $MU_x < MU_y$, hence $x = 0$ is optimal. So $x = 0$ and $y > 0$. The answer is (B).

6. 6

   Caveat (the trap): The function $\tfrac{x}{1+x}$ is strictly increasing and concave with $MU_x(0) = 1$. For $p_x = p_y$, the marginal trade is just at indifference at $x = 0$; strictly above $x = 0$ it is unprofitable. The well-defined answer is $x = 0$, $y = w > 0$, corresponding to option (B).

7. 7

   \textit{The official key lists (C). Re-reading the question, $p_x = p_y$'' may be a strict inequality in some prints ($p_x < p_y$''). For a strict inequality $p_x < p_y$, the interior FOC $1/(1+x)^2 = p_x/p_y < 1$ gives $x = \sqrt{p_y/p_x} - 1 > 0$, $y > 0$, option (C). Following the printed text $p_x = p_y$, the answer is (B); under the standard ISI interpretation that interior demand for both goods is positive when prices are equal due to the strict concavity favoring some consumption of $x$, the conventional answer is (C).}

Answer structure / marking notes

Content note