Cost Curves

This scene pulls marginal and average cost curves out of a total cost curve.

Marginal cost is the extra cost of making one more unit. Average cost is total cost divided by quantity. When one more unit costs less than the average, the average falls. So the marginal cost curve passes exactly through the lowest point of average cost.

For example, a 🏪 shop's average cost is 1,000 won per 🍞 bread loaf. Baking one more costs 800 won, so the average falls. At 1,200 won, the average rises instead. That is why marginal cost meets average cost at its lowest point.

C(q)=F+aq+bq2+cq3C(q) = F + aq + bq^2 + cq^3
MC(q)=C′(q),AC(q)=C(q)qMC(q) = C'(q), \quad AC(q) = \frac{C(q)}{q}

Part of these learning paths

Examples & Exercises

For C(q)=40+30q−4q2+12q3C(q) = 40 + 30q - 4q^2 + \tfrac{1}{2}q^3, find the output minimizing average cost and show that MC=ACMC = AC there.

AC(q)=40q+30−4q+12q2,AC′(q)=−40q2−4+q=0AC(q) = \frac{40}{q} + 30 - 4q + \tfrac{1}{2}q^2, \quad AC'(q) = -\frac{40}{q^2} - 4 + q = 0
q∗≈5.38,AC(q∗)≈30.39=MC(q∗)q^* \approx 5.38, \quad AC(q^*) \approx 30.39 = MC(q^*)
Cubic cost — marginal cost pierces the average-cost minimum.
Show proof

  1. AC′(q)=−40/q2−4+qAC'(q) = -40/q^2 - 4 + q. It equals −1.6−4+5=−0.6<0-1.6 - 4 + 5 = -0.6 < 0 at q=5q = 5 and −1.11−4+6=0.89>0-1.11 - 4 + 6 = 0.89 > 0 at q=6q = 6, so the root lies in (5,6)(5, 6); numerically q∗≈5.38q^* \approx 5.38.

  2. AC(5.38)=40/5.38+30−4(5.38)+0.5(5.38)2≈30.39AC(5.38) = 40/5.38 + 30 - 4(5.38) + 0.5(5.38)^2 \approx 30.39. Plugging into marginal cost MC(q)=30−8q+1.5q2MC(q) = 30 - 8q + 1.5q^2 gives MC(5.38)≈30.39MC(5.38) \approx 30.39 — they agree.

  3. No coincidence: by the quotient rule AC′(q)=(MC(q)⋅q−C(q))/q2AC'(q) = (MC(q)\cdot q - C(q))/q^2, so AC′(q)=0  ⟺  MC(q)=AC(q)AC'(q) = 0 \iff MC(q) = AC(q). The marginal cost curve passes exactly through every extremum of the average cost curve.

Show that fixed cost FF does not affect marginal cost, and that with F=0F = 0 we have AC=AVCAC = AVC with coinciding minima.

MC(q)=ddq(F+aq+bq2+cq3)=a+2bq+3cq2MC(q) = \frac{d}{dq}(F + aq + bq^2 + cq^3) = a + 2bq + 3cq^2
F=0  ⇒  AC(q)=AVC(q)=30−4q+12q2,min⁡ at q=4,  =22F = 0 \;\Rightarrow\; AC(q) = AVC(q) = 30 - 4q + \tfrac{1}{2}q^2, \quad \min \text{ at } q = 4,\; = 22
F=0F = 0 — average and average-variable cost coincide.
Show proof

  1. Differentiating C(q)C(q) kills the constant FF, so MC(q)=a+2bq+3cq2MC(q) = a + 2bq + 3cq^2 is independent of it. Fixed cost does not affect marginal decisions — it is sunk.

  2. With F=0F = 0, AC(q)=(30q−4q2+0.5q3)/q=30−4q+0.5q2=AVC(q)AC(q) = (30q - 4q^2 + 0.5q^3)/q = 30 - 4q + 0.5q^2 = AVC(q). From dAC/dq=−4+q=0dAC/dq = -4 + q = 0, q=4q = 4 with value 30−16+8=2230 - 16 + 8 = 22, and MC(4)=30−32+24=22MC(4) = 30 - 32 + 24 = 22 — MC pierces the shared minimum.

6. Production