PEA 2022 PEA Q10 | ISI MSQE

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PEAMCQModerate

2022 PEA Q10

$F(K,L) = \min\{aK, bL\}$ , $a,b>0$ , $a \neq b$ . For fixed $\bar K$ , the marginal product of labor is

Reveal answer and solution

Answer

Solution

1
For fixed $\bar K$ ,
2
$F(\bar K, L) = \begin{cases} bL & \text{if } bL < a\bar K,\ \text{i.e.}\ L < (a/b)\bar K, \\ a\bar K & \text{if } L \geq (a/b)\bar K. \end{cases}$
3
Hence
4
$\text{MP}_L = \begin{cases} b & \text{if } L < (a/b)\bar K, \\ 0 & \text{if } L > (a/b)\bar K. \end{cases}$
5
Options (B) and (C) have the wrong coefficient ( $1/a$ or $1/b$ instead of $b$ ) and/or the wrong threshold. Therefore none of the listed options is correct.

Answer structure / marking notes

In $\min\{aK, bL\}$ , the marginal product of labor (when labor is the binding factor) equals the coefficient on $L$ , namely $b$ , not $1/b$ .

Content note

Imported from public/resources/isi/msqe/solutions/pea/2022/ISI_MSQE_PEA_2022_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.