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Worked Examples · Example 2

Q.Using the same schedule as Worked Example 1, state at which level of labour Average Product is maximum, and verify the relationship between Marginal Product and Average Product at that point.

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✓ Free question

Given: the AP and MP columns computed in Worked Example 1.

Step 1 — Locate the maximum AP: the AP values are 10,12,13,13,12,10,810, 12, 13, 13, 12, 10, 8. The highest value is 1313, reached at L=3L=3 and L=4L=4; the last level at which AP is at its peak before it starts to fall is L=4L=4.

Step 2 — Read MP at that point: at L=4L=4, MP=13MP=13, which equals AP=13AP=13.

Step 3 — Check the relationship on both sides:

  • Before the peak, at L=3L=3: MP=15MP=15, AP=13AP=13, so MP>APMP>AP — and indeed AP was still rising (from 13 towards its peak).
  • At the peak, L=4L=4: MP=AP=13MP=AP=13.
  • After the peak, at L=5L=5: MP=8MP=8, AP=12AP=12, so MP<APMP<AP — and indeed AP is now falling (from 13 to 12).

This confirms the general rule: AP rises while MP>APMP>AP, is maximum where MP=APMP=AP, and falls while MP<APMP<AP.

Check (independent method — averaging logic): average product is a running average of the marginal products. A new marginal value above the current average must pull the average up; one below it pulls the average down; the two are equal exactly at the turning point of the average — which is why MP=APMP=AP coincides with the maximum of AP, matching the table.

✓Final answer

AP is maximum (13) at L=4L=4, where MP=AP=13MP=AP=13; MP lies above AP while AP rises and below AP once AP falls.

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