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2025 PEA Q2

Continue with Q1. Assume δ=0.1\delta = 0.1. At the steady state, consumption per worker and the saving rate that maximises consumption per worker are given by, respectively:

Reveal answer and solution

Answer

A

Solution

  1. 1

    From Q1, output per worker yˉ=s/δ\bar y = s/\delta and steady-state consumption per worker is

  2. 2
    cˉ=(1−s)yˉ=(1−s)sδ. \bar c = (1-s)\bar y = (1-s)\frac{s}{\delta}.
  3. 3

    Substituting δ=0.1\delta = 0.1:

  4. 4
    cˉ=s(1−s)0.1=10 s(1−s). \bar c = \frac{s(1-s)}{0.1} = 10\, s(1-s).
  5. 5

    Up to the normalisation, cˉ∝s(1−s)\bar c \propto s(1-s). To maximise,

  6. 6
    dcˉds∝1−2s=0  ⟹  s∗=0.5. \frac{d\bar c}{ds} \propto 1 - 2s = 0 \;\Longrightarrow\; s^* = 0.5.

Answer structure / marking notes

The golden rule here gives s∗=0.5s^* = 0.5 because the per-worker production is effectively linear (after normalisation), so the maximisation reduces to a simple quadratic in ss.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.