Parcel-Sorting Example: MC Crossing AC

Scenario: A small parcel-sorting station has a fixed equipment cost of €4 per shift and a basic variable handling cost of €1 per parcel. Up to 4 parcels, the station works efficiently. Beyond 4 parcels, congestion and overtime increase the extra cost of each additional parcel.

Equations

FC = 4

VC(q) = q + 0.5 [max(0, q − 4)]²

TC(q) = FC + VC(q)

AC(q) = TC(q) / q

Continuous MC(q) = dTC/dq = 1 (q ≤ 4); q − 3 (q > 4)

Discrete MC(q) = TC(q) − TC(q − 1) (whole parcels)

Why are the two MC values different at q = 5? The whole fifth parcel adds €9.50 − €8.00 = €1.50 to total cost. But the instantaneous slope at q = 5 is €2.00. The red curve uses the instantaneous derivative, so it intersects AC at the continuous minimum.

The two approaches describe different quantities: a finite increase of one parcel versus a slope at a point.

Intersection / minimum AC point:

Table

q	FC	VC	TC	AFC	AVC	AC	MC (ΔTC)	MC (derivative)

Chart

Average Cost (AC) Continuous Marginal Cost (dTC/dq) Minimum AC / intersection point

The red line is instantaneous marginal cost, not the cost of each whole additional parcel. For integer parcel counts, use the MC (ΔTC) column in the table. The AC curve reaches its continuous minimum around q = 4.90; the best whole-number quantity is 5 parcels.