Cost Functions

Applied EconomicsDifficulty: ███░░Depth: 7Unlocks: 6

Fixed, variable, and marginal cost. Average cost curves, economies of scale. Short-run vs long-run cost structures.

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Understanding cost functions tells you when a firm should expand, contract, or build a new plant — and why prices sometimes fall when firms get bigger.

TL;DR:

Cost functions describe how total, fixed, variable, average, and marginal costs depend on output; they let you use calculus (Derivatives) and optimization to find profit-maximizing or cost-minimizing output and evaluate economies of scale.

What Is Cost Functions?

Cost functions are mathematical descriptions of how a firm's costs depend on the quantity of output it produces. At its most basic, a cost function answers: if the firm produces qq units, what is the total cost TC(q)TC(q)? Splitting total cost into components yields intuition and decision rules that managers and economists use.

Key pieces and why they matter

  • Total Cost (TC(q)TC(q)): The total economic cost of producing qq units, usually measured in dollars. Example: if producing nothing still requires some expenses (like rent), then TC(0)>0TC(0)>0. Concretely, consider a simple quadratic cost function
TC(q)=100+5q+0.5q2.TC(q)=100+5q+0.5q^2.

Numerically, TC(0)=100TC(0)=100 (the firm has a 100fixedcostifitshutsdownoutput),and100 fixed cost if it shuts down output), and TC(10)=100+5(10)+0.5(10)^2=100+50+50=200$.

  • Fixed Cost (FCFC): Costs that do not vary with qq in the short run (e.g., rent, machinery that must be paid for whether you use it or not). In the example, FC=100FC=100. Fixed cost is the vertical intercept of TC(q)TC(q).
  • Variable Cost (VC(q)VC(q)): Costs that vary with output (e.g., raw materials, hourly labor). By definition VC(q)=TC(q)FCVC(q)=TC(q)-FC. In the example,
VC(q)=5q+0.5q2.VC(q)=5q+0.5q^2.

For q=10q=10, VC(10)=5(10)+0.5(100)=50+50=100VC(10)=5(10)+0.5(100)=50+50=100.

  • Average Cost measures: These let you compare per-unit costs across output levels.
  • Average Total Cost (ATC or ACAC): AC(q)=TC(q)qAC(q)=\dfrac{TC(q)}{q}. For the example,
AC(q)=100+5q+0.5q2q=100q+5+0.5q.AC(q)=\frac{100+5q+0.5q^2}{q}=\frac{100}{q}+5+0.5q.

If q=10q=10, AC(10)=10010+5+0.5(10)=10+5+5=20AC(10)=\frac{100}{10}+5+0.5(10)=10+5+5=20.

  • Average Fixed Cost (AFCAFC): AFC(q)=FCq=100qAFC(q)=\dfrac{FC}{q}=\dfrac{100}{q}. For q=10q=10, AFC(10)=10AFC(10)=10.
  • Average Variable Cost (AVCAVC): AVC(q)=VC(q)q=5+0.5qAVC(q)=\dfrac{VC(q)}{q}=5+0.5q. For q=10q=10, AVC(10)=5+5=10AVC(10)=5+5=10.

Notice AC(q)=AFC(q)+AVC(q)AC(q)=AFC(q)+AVC(q) numerically: 20=10+1020=10+10.

  • Marginal Cost (MCMC): The instantaneous additional cost of producing one more unit. In calculus terms (see Derivatives (d2)), marginal cost is the derivative of total cost with respect to qq:
MC(q)=TC(q).MC(q)=TC'(q).

For our example,

MC(q)=ddq(100+5q+0.5q2)=5+q.MC(q)=\frac{d}{dq}\left(100+5q+0.5q^2\right)=5+q.

Numerically, MC(10)=5+10=15MC(10)=5+10=15. Marginal cost approximates the cost of the 11th unit; using discrete difference, TC(11)TC(10)=(100+5(11)+0.5(11)2)200=100+55+60.5200=15.5TC(11)-TC(10)=\big(100+5(11)+0.5(11)^2\big)-200=100+55+60.5-200=15.5, close to MC(10)=15MC(10)=15 since the function is smooth.

Short-run vs Long-run

  • Short-run: At least one input is fixed (hence fixed cost exists). The short-run cost functions (FC, VC, MC, AC) are conditional on the current plant size or capacity.
  • Long-run: All inputs are variable. There is no fixed cost in the long run for the firm's planning problem. Long-run cost (LRTC or LACLAC) is the envelope of short-run average costs across all possible plant sizes. Long-run cost determines optimal scale, which leads to the concept of economies of scale.

Economies of scale

  • If LAC(q)LAC(q) decreases as qq increases over some range, the firm enjoys economies of scale (average cost per unit falls with output). If LAC(q)LAC(q) increases with qq, there are diseconomies of scale. If LAC(q)LAC(q) is flat, constant returns to scale.

Why this matters: Marginal cost interacts with price and marginal revenue to determine optimal output (see Optimization Introduction (d3)). Average cost tells you whether a firm covers its costs at a given price and whether expanding output reduces unit costs.

Core Mechanic 1: Marginal and Average Costs — Calculus Links and Local Decision Rules

Marginal cost (MC) is central because in competitive markets the profit-maximizing condition often equates price to marginal cost (when firms are price takers). Average costs determine profitability and long-run entry/exit.

Formal relationships and calculus intuition (reference: Derivatives (d2))

  • Marginal cost as a derivative: Given TC(q)TC(q) differentiable,
MC(q)=TC(q).MC(q)=TC'(q).

Example: If TC(q)=50+2q+0.1q2TC(q)=50+2q+0.1q^2, then

MC(q)=2+0.2q.MC(q)=2+0.2q.

Numerically, MC(20)=2+0.2(20)=2+4=6MC(20)=2+0.2(20)=2+4=6. So the additional cost at output 20 is $6 per extra unit.

  • Marginal cost vs average cost slope: A basic calculus result (related to the derivative of a quotient) tells us that MC(q)MC(q) intersects AC(q)AC(q) at the ACAC minimum. Consider AC(q)=TC(q)/qAC(q)=TC(q)/q. Differentiate using the quotient rule (or product rule if you write AC(q)=TC(q)q1AC(q)=TC(q)\cdot q^{-1}):
AC(q)=qTC(q)TC(q)q2=qMC(q)TC(q)q2.AC'(q)=\frac{q\cdot TC'(q)-TC(q)}{q^2}=\frac{q\cdot MC(q)-TC(q)}{q^2}.

Setting AC(q)=0AC'(q)=0 gives qMC(q)TC(q)=0MC(q)=TC(q)/q=AC(q)q\cdot MC(q)-TC(q)=0\Rightarrow MC(q)=TC(q)/q=AC(q). Thus the stationary point of ACAC occurs where MC=ACMC=AC. If that point is a minimum, MC crosses AC from below.

Concrete numeric check: Take TC(q)=100+5q+0.5q2TC(q)=100+5q+0.5q^2 as before. Then AC(q)=100/q+5+0.5qAC(q)=100/q+5+0.5q and MC(q)=5+qMC(q)=5+q. Solve MC(q)=AC(q)MC(q)=AC(q):

5+q=100q+5+0.5qq=100q+0.5qq0.5q=100q0.5q=100qq2=200q=20014.142.5+q=\frac{100}{q}+5+0.5q\Rightarrow q=\frac{100}{q}+0.5q\Rightarrow q-0.5q=\frac{100}{q}\Rightarrow 0.5q=\frac{100}{q}\Rightarrow q^2=200\Rightarrow q=\sqrt{200}\approx14.142.

Numerically check: MC(14.142)5+14.142=19.142MC(14.142)\approx5+14.142=19.142. And AC(14.142)=100/14.142+5+0.5(14.142)7.071+5+7.071=19.142AC(14.142)=100/14.142+5+0.5(14.142)\approx7.071+5+7.071=19.142, matching.

  • Behavior of MC and AC curves: Because AFCAFC falls with qq (since AFC=FC/qAFC=FC/q), ACAC initially falls if AVCAVC is relatively flat and AFCAFC dominates. But if MCMC is increasing (e.g., due to diminishing marginal returns), MCMC eventually pulls ACAC up. Graphically, MCMC typically has a U-shape and crosses the U-shaped ACAC curve at its minimum.
  • Marginal cost from variable cost: Since FCFC is constant, MC(q)=TC(q)=VC(q)MC(q)=TC'(q)=VC'(q). So for calculus problems it's often enough to specify VC(q)VC(q). Example: If VC(q)=20q+0.4q2VC(q)=20q+0.4q^2, then MC(q)=20+0.8qMC(q)=20+0.8q. For q=25q=25, MC(25)=20+0.8(25)=20+20=40MC(25)=20+0.8(25)=20+20=40.

Local decision rules and short-run shutdown condition (link to Optimization Introduction (d3))

  • Profit or loss short-run: Profit is π(q)=TR(q)TC(q)\pi(q)=TR(q)-TC(q) where TR(q)=pqTR(q)=p\cdot q in perfect competition. First-order condition (from Optimization Introduction (d3) and derivatives) for an interior optimum: π(q)=0pTC(q)=0p=MC(q)\pi'(q)=0\Rightarrow p-TC'(q)=0\Rightarrow p=MC(q).

Example numeric decision: If price p=19p=19 and MC(q)=5+qMC(q)=5+q, then optimal output solves 5+q=19q=145+q=19\Rightarrow q=14. This matches the earlier ACAC minimum example and shows how calculus directly produces the optimal output.

  • Shutdown rule: In the short run, if price pp is below the minimum of AVC(q)AVC(q), the firm should shut down and produce q=0q=0 (it would minimize losses by paying fixed costs only). If pp exceeds minimum AVCAVC, produce where p=MCp=MC.

Concrete numbers: For VC(q)=5q+0.5q2VC(q)=5q+0.5q^2, AVC(q)=5+0.5qAVC(q)=5+0.5q has minimum at q=0q=0 (since it's increasing here) with AVC(0)=5AVC(0)=5. If market price p=4p=4, since p<5p<5 the firm should shut down. If p=10p=10, solve MC=5+q=10q=5MC=5+q=10\Rightarrow q=5.

This section shows how derivatives (Derivatives (d2)) and optimization first-order conditions (Optimization Introduction (d3)) translate to managerial rules: set output where p=MCp=MC, check AVCAVC to decide shutdown, and use ACAC to check profitability.

Core Mechanic 2: Short-Run vs Long-Run Costs and Economies of Scale

Understanding the distinction between short-run and long-run cost structures is essential for planning capacity and for understanding industry dynamics (entry, exit, and firm size).

Short-run cost structure

  • In the short run, at least one input is fixed (e.g., capital). The firm faces a specific TCSR(q)=FC+VCSR(q)TC_{SR}(q)=FC+VC_{SR}(q) for its current plant. Fixed cost FCFC does not change with qq in the short run. Consequently, AFC(q)=FC/qAFC(q)=FC/q declines as qq rises.

Example: Suppose a firm has a factory requiring FC=200FC=200 and variable cost VCSR(q)=10q+0.2q2VC_{SR}(q)=10q+0.2q^2. Then

TCSR(q)=200+10q+0.2q2,TC_{SR}(q)=200+10q+0.2q^2,
MCSR(q)=10+0.4q,MC_{SR}(q)=10+0.4q,
ACSR(q)=200q+10+0.2q.AC_{SR}(q)=\frac{200}{q}+10+0.2q.

Numerical check: at q=20q=20, TCSR(20)=200+200+80=480TC_{SR}(20)=200+200+80=480 (wait: compute carefully: 1020=20010*20=200, 0.2400=800.2*400=80, so TCSR=480TC_{SR}=480), MCSR(20)=10+0.4(20)=18MC_{SR}(20)=10+0.4(20)=18, ACSR(20)=200/20+10+0.2(20)=10+10+4=24AC_{SR}(20)=200/20+10+0.2(20)=10+10+4=24.

  • The short-run average cost curve for a given plant is typically U-shaped because AFCAFC falls with qq (pulling ACAC down at low q) while rising marginal costs cause AVCAVC to rise and eventually pull ACAC up.

Long-run cost structure and the envelope theorem

  • Long-run costs allow adjustment of all inputs. For each target output qq, the firm can choose the plant size or input mix that minimizes cost. Long-run total cost LTC(q)LTC(q) equals the minimal short-run total cost across all feasible plant sizes. Graphically, the long-run average cost curve LAC(q)LAC(q) is the lower envelope of the short-run average cost curves for different plant sizes.

Algebraically, suppose plant size is parameterized by kk (e.g., capacity). For each kk we have TC(q;k)=FC(k)+VC(q;k)TC(q;k)=FC(k)+VC(q;k). Then

LTC(q)=minkTC(q;k).LTC(q)=\min_k TC(q;k).

Example (simplified): Suppose there are two discrete plant sizes:

  • Small plant: TCS(q)=100+6q+0.2q2TC_S(q)=100+6q+0.2q^2 (good for low q)
  • Large plant: TCL(q)=400+3q+0.1q2TC_L(q)=400+3q+0.1q^2 (more fixed cost but lower variable cost)

For q=10q=10, TCS(10)=100+60+20=180TC_S(10)=100+60+20=180, TCL(10)=400+30+10=440TC_L(10)=400+30+10=440. So the small plant is cheaper at low output. For q=200q=200, TCS(200)=100+1200+8000=9320TC_S(200)=100+1200+8000=9320, TCL(200)=400+600+4000=5000TC_L(200)=400+600+4000=5000; the large plant is cheaper at high output. The long-run cost at each qq picks the cheaper plant, creating an LAC(q)LAC(q) that can fall then rise depending on technology.

Economies of scale (technical/market meanings)

  • Economies of scale: If LAC(q)LAC(q) decreases as qq increases over some range, doubling output less than doubles cost (average cost falls). For example, if LTC(q)=aqαLTC(q)=aq^\alpha with α<1\alpha<1, then LAC(q)=aqα1LAC(q)=aq^{\alpha-1} falls with qq.

Concrete numeric instance: Take LTC(q)=10q0.8LTC(q)=10q^{0.8}. Then LAC(q)=10q0.2LAC(q)=10q^{-0.2}. For q=100q=100, LAC(100)=10(100)0.2=10/(1000.2)=10/(2.5119)3.98LAC(100)=10(100)^{-0.2}=10/(100^{0.2})=10/(\approx2.5119)\approx3.98. For q=1000q=1000, LAC(1000)=10(1000)0.2=10/(10000.2)=10/(3.981)2.51LAC(1000)=10(1000)^{-0.2}=10/(1000^{0.2})=10/(\approx3.981)\approx2.51. So average cost falls with scale: economies of scale.

  • Sources of economies of scale: specialization, indivisibilities in capital (a huge machine that spreads cost across many units), bulk buying of inputs, or spreading fixed costs. Sources of diseconomies: managerial coordination costs, congestion, or increasing input prices when scaling.

Returns to scale vs economies of scale distinction

  • Returns to scale (a production function concept) measures how output responds when all inputs scale. If doubling inputs more than doubles output, there are increasing returns to scale. This translates into cost language: increasing returns to scale in production typically imply economies of scale (falling LACLAC), but care is needed because input prices and discrete plant options can alter the mapping.

Long-run planning decisions

  • Firms choose the plant size to minimize average cost for their expected output in the long run. If demand grows, a firm may invest in a larger plant to move to a lower LACLAC region.
  • Industry implications: If LACLAC falls over a wide range such that large firms have much lower LACLAC than small ones, the industry is natural monopoly-prone (one large firm can supply at lower average cost). If LACLAC is fairly flat, many firms can coexist.

Applications and Connections

Cost functions are used across microeconomics, managerial economics, and public policy. Here are concrete applications and how this connects to downstream topics.

Pricing and output decisions in market structures

  • Perfect competition: Firms are price takers. Profit maximization uses calculus (Optimization Introduction (d3)): set p=MC(q)p=MC(q) if pAVCminp\geq AVC_{min}; otherwise shut down. Example: If market price p=30p=30 and MC(q)=10+0.5qMC(q)=10+0.5q, solve 10+0.5q=30q=4010+0.5q=30\Rightarrow q=40. If AVCmin=12AVC_{min}=12 and p=11p=11, firm should shut down.
  • Monopoly/monopolistic competition: Marginal cost still matters, but the profit-maximizing rule uses marginal revenue MR(q)MR(q): set MR(q)=MC(q)MR(q)=MC(q). For a linear demand P(q)=1002qP(q)=100-2q, TR(q)=100q2q2TR(q)=100q-2q^2 so MR(q)=1004qMR(q)=100-4q. With MC(q)=20+qMC(q)=20+q, the optimum solves 1004q=20+q5q=80q=16100-4q=20+q\Rightarrow5q=80\Rightarrow q=16, charging P(16)=10032=68P(16)=100-32=68.

Cost-benefit, entry/exit, and long-run industry supply

  • Entry and exit: In the long run, firms enter if P>LAC(q)P>LAC(q^*) for potential entrants and exit if P<LAC(q)P<LAC(q^*) for incumbents. For example, if the competitive market price converges to the minimum of LACLAC, firms earn zero economic profit in a free-entry long-run equilibrium.
  • Natural monopoly and regulation: If LACLAC declines across the entire relevant output range (due to large fixed costs), a single firm produces at lower cost than many small ones. Regulators may set price equal to ACAC to allow zero economic profit, or to MCMC, sometimes with subsidies if MC<ACMC<AC for all q.

Cost estimation and managerial accounting

  • Empirical cost estimation fits functional forms (linear, quadratic, translog) to observed costs. The marginal cost derived from estimated TC(q)TC(q) then informs pricing decisions. Example: if regression yields TC(q)=150+8q+0.3q2TC(q)=150+8q+0.3q^2, managers compute MC(q)=8+0.6qMC(q)=8+0.6q to set incremental pricing.

Network industries and scale economies

  • In software or platform firms, marginal cost of an additional customer can be near zero (digital goods). Here MC0MC\approx0 and average cost falls dramatically with output (huge economies of scale). Pricing strategies (freemium, subscription) exploit low marginal cost and high fixed development cost.

Connection to welfare analysis

  • In public policy, comparing PP to MCMC is central for efficiency: price equals marginal cost is allocatively efficient. When MCMC is below ACAC due to high fixed costs, setting price at MCMC can require subsidy; regulators weigh efficiency vs distribution.

Downstream topics that use cost function mechanics

  • Production theory: returns to scale and cost functions are duals; cost minimization problems produce cost functions and conditional factor demands.
  • Firm dynamics and industrial organization: long-run cost shapes market structure (number and scale of firms). Empirical IO models estimate LACLAC to understand barriers to entry.
  • Game theory and strategic pricing: knowledge of rival cost curves informs price competition (Bertrand) and quantity competition (Cournot).

Concrete managerial checklist

  • Compute MC(q)=TC(q)MC(q)=TC'(q) and AC(q)=TC(q)/qAC(q)=TC(q)/q for candidate qq.
  • If price is given, check shutdown: if p<minAVCp<\min AVC, produce q=0q=0; otherwise choose qq with p=MC(q)p=MC(q).
  • For long-run planning, compute LAC(q)LAC(q) by minimizing TC(q;k)TC(q;k) over plant choices; identify range of qq where LACLAC falls (economies of scale).

Every formula in this lesson ties back to calculus (Derivatives (d2)) for rates of change and to optimization first-order conditions (Optimization Introduction (d3)) for choosing qq to maximize profit or minimize cost. These tools give actionable rules for pricing, capacity choice, and policy design.

Worked Examples (3)

Basic MC and AC computation

Given TC(q)=120+4q+0.25q2TC(q)=120+4q+0.25q^2, compute FCFC, VC(q)VC(q), MC(q)MC(q), AC(q)AC(q), and evaluate them at q=40q=40.

  1. Identify fixed cost (constant term): FC=120FC=120.

  2. Compute variable cost: VC(q)=TC(q)FC=4q+0.25q2VC(q)=TC(q)-FC=4q+0.25q^2. For q=40q=40, VC(40)=4(40)+0.25(1600)=160+400=560VC(40)=4(40)+0.25(1600)=160+400=560.

  3. Differentiate to get marginal cost: MC(q)=TC(q)=4+0.5qMC(q)=TC'(q)=4+0.5q. For q=40q=40, MC(40)=4+0.5(40)=4+20=24MC(40)=4+0.5(40)=4+20=24.

  4. Compute average total cost: AC(q)=TC(q)/q=120+4q+0.25q2q=120q+4+0.25qAC(q)=TC(q)/q=\frac{120+4q+0.25q^2}{q}=\frac{120}{q}+4+0.25q. For q=40q=40, AC(40)=120/40+4+0.25(40)=3+4+10=17AC(40)=120/40+4+0.25(40)=3+4+10=17.

  5. Verify AC=AVC+AFCAC=AVC+AFC: AFC(40)=120/40=3AFC(40)=120/40=3, AVC(40)=VC(40)/40=560/40=14AVC(40)=VC(40)/40=560/40=14, sum 3+14=173+14=17, matching AC(40)AC(40).

Insight: This example reinforces that MCMC is the derivative of TCTC, FCFC is the constant term, and averages decompose cleanly; numerically checking identities is a useful verification step.

Shutdown decision using AVC minimum

A firm has TC(q)=200+8q+0.5q2TC(q)=200+8q+0.5q^2. Market price is p=9p=9. Should the firm produce in the short run? If yes, find optimal qq; if no, explain why.

  1. Compute VC(q)=8q+0.5q2VC(q)=8q+0.5q^2 and AVC(q)=VC(q)q=8+0.5qAVC(q)=\frac{VC(q)}{q}=8+0.5q.

  2. Find minimum of AVC(q)AVC(q): AVC is linear increasing in qq here with slope $0.5$, so the minimum occurs at q=0q=0 with AVC(0)=8AVC(0)=8. (Interpretation: marginally, AVC increases from 8.)

  3. Compare market price p=9p=9 to minAVC=8\min AVC=8. Since p>8p>8, the firm should produce (not shut down) because price covers variable cost at some positive output.

  4. Compute marginal cost: MC(q)=TC(q)=8+qMC(q)=TC'(q)=8+q. Set p=MCp=MC to find interior optimum: 9=8+qq=19=8+q\Rightarrow q=1.

  5. Compute profit (optional check): TR=9(1)=9TR=9(1)=9, TC(1)=200+8+0.5=208.5TC(1)=200+8+0.5=208.5, profit =9208.5=199.5=9-208.5=-199.5. Loss is large, but since loss (199.5)(-199.5) is less than fixed cost loss of 200-200 from shutting down, producing minimizes loss in the short run.

Insight: Even when producing yields an economic loss, producing can be better than shutting down if price covers variable costs; calculus provides the production rule p=MCp=MC for interior solutions.

Long-run plant choice and economies of scale

Two plant options: Small plant TCS(q)=150+6q+0.4q2TC_S(q)=150+6q+0.4q^2, Large plant TCL(q)=500+3q+0.2q2TC_L(q)=500+3q+0.2q^2. For q=50q=50 determine which plant minimizes cost. Then compute LAC(50)LAC(50) and comment on economies of scale comparing q=50q=50 to q=200q=200.

  1. Compute TCS(50)=150+6(50)+0.4(2500)=150+300+1000=1450TC_S(50)=150+6(50)+0.4(2500)=150+300+1000=1450.

  2. Compute TCL(50)=500+3(50)+0.2(2500)=500+150+500=1150TC_L(50)=500+3(50)+0.2(2500)=500+150+500=1150.

  3. Compare costs: TCL(50)=1150<TCS(50)=1450TC_L(50)=1150<TC_S(50)=1450, so the large plant is cheaper for q=50q=50.

  4. Compute long-run average cost at q=50q=50: LAC(50)=TCL(50)/50=1150/50=23LAC(50)=TC_L(50)/50=1150/50=23 (since large plant chosen).

  5. Now evaluate at q=200q=200: TCS(200)=150+1200+0.4(40000)=150+1200+16000=17350TC_S(200)=150+1200+0.4(40000)=150+1200+16000=17350. TCL(200)=500+600+0.2(40000)=500+600+8000=9100TC_L(200)=500+600+0.2(40000)=500+600+8000=9100. Then LAC(200)=9100/200=45.5.CompareLAC(200)=9100/200=45.5. Compare LAC(50)=23to to LAC(200)=45.5$: average cost increased going from 50 to 200, indicating diseconomies of scale in this numeric example across that range. (Interpret: the large plant is better at low to medium q here but both plants generate higher average cost at much larger q due to quadratic variation.)

Insight: Picking among plant options is an optimization over discrete or continuous capital choices; the long-run average cost is the lower envelope of short-run costs and can exhibit economies or diseconomies of scale depending on technology parameters.

Key Takeaways

  • Total cost decomposes into fixed cost (doesn't vary with q in the short run) and variable cost (does); numerically compute FC as the constant term in polynomial TC(q).

  • Marginal cost equals the derivative of total cost: MC(q)=TC(q)MC(q)=TC'(q) (Derivatives (d2)); in competitive settings set p=MCp=MC for interior optima (Optimization Introduction (d3)).

  • Average cost equals total cost per unit: AC(q)=TC(q)/qAC(q)=TC(q)/q, and MCMC intersects ACAC at ACAC's minimum point.

  • Short-run cost curves reflect at least one fixed input; long-run cost minimizes over plant sizes and is the envelope of short-run curves.

  • Economies of scale mean declining long-run average cost with output; this can justify large-scale firms, but diseconomies can arise at large sizes.

  • Shutdown rule: produce in short run iff pminAVC(q)p\geq\min AVC(q); otherwise shut down and accept fixed costs without producing.

  • Empirical and managerial uses: estimate TC to compute MC for pricing, choose plant size to minimize LAC, and assess industry structure (natural monopoly vs competitive).

Common Mistakes

  • Confusing marginal cost with average cost: MC is the derivative (slope of TC), not TC divided by q. This leads to wrong decisions—MC can be below AC while AC is falling, but they are equal only at AC's minimum.

  • Treating fixed cost as avoidable in the short run: FC is sunk in the short run and should not affect marginal production decisions, only long-run planning. Incorporating FC into the shutdown decision is incorrect.

  • Using discrete differences instead of derivatives without checking scale: for rapidly changing or non-smooth TC functions, the discrete increment TC(q+1)TC(q)TC(q+1)-TC(q) can diverge from TC(q)TC'(q); always check the function curvature.

  • Assuming long-run average cost equals short-run average cost for a single plant: long-run cost optimizes over plant sizes, so LACLAC is the lower envelope of SR averages; using a single SR curve for long-run decisions can mislead investment choices.

Practice

easy

Easy: Given TC(q)=80+3q+0.1q2TC(q)=80+3q+0.1q^2, compute MC(q)MC(q) and AC(q)AC(q), then evaluate MC(20)MC(20) and AC(20)AC(20).

Hint: Differentiate TCTC for MCMC. For ACAC, divide TCTC by qq and plug in q=20q=20.

Show solution

MC(q)=3+0.2q, so MC(20)=3+0.2(20)=7. AC(q)=80/q+3+0.1q, so AC(20)=80/20+3+0.1(20)=4+3+2=9.

medium

Medium: A firm has short-run cost TCSR(q)=250+10q+0.5q2TC_{SR}(q)=250+10q+0.5q^2. Market price is p=35p=35. Should the firm produce in the short run? If yes, find the profit-maximizing qq and the resulting profit.

Hint: Compute AVCAVC and its minimum to check shutdown; then set p=MCp=MC for the interior solution (Optimization Introduction (d3)).

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VC(q)=10q+0.5q^2 so AVC(q)=10+0.5q, minimum at q=0 with AVC_min=10. Since p=35>10, produce. MC(q)=10+q. Set p=MC: 35=10+q => q=25. Compute TR=3525=875. TC=250+1025+0.5*625=250+250+312.5=812.5. Profit=875-812.5=62.5.

hard

Hard: You are choosing plant size k≥0 affecting fixed and variable costs: FC(k)=100kFC(k)=100k, VC(q;k)=5q+0.2k1q2VC(q;k)=5q+0.2k^{-1}q^2. For a required output q0=100, choose k to minimize TC(q0;k)=100k+5(100)+0.2k1(100)2TC(q_0;k)=100k+5(100)+0.2k^{-1}(100)^2. Find optimal k and the minimized total and average cost. (Edge case reasoning: ensure k>0.)

Hint: Treat TC as a function of k and differentiate w.r.t k; use first-order condition and check second derivative for minimization. This uses Optimization Introduction (d3).

Show solution

Write TC(k)=100k+500+0.2(10000)/k=100k+500+2000/k. Differentiate: TC'(k)=100-2000/k^2. Set =0 => 100=2000/k^2 => k^2=2000/100=20 => k=\sqrt{20}\approx4.4721 (positive root). Second derivative TC''(k)=4000/k^3>0 so minimizer. Minimized TC=100(4.4721)+500+2000/4.4721 ≈447.21+500+447.21=1394.42. Average cost =TC/100 ≈13.9442.

Connections

Looking back: In Derivatives (d2) we learned that derivatives give instantaneous rates of change; here MC(q)=TC(q)MC(q)=TC'(q) is a direct application of that concept to economics. In Optimization Introduction (d3) we learned how to set first-order conditions to find maxima/minima under constraints; here we set p=MCp=MC (or MR=MCMR=MC) and differentiate cost with respect to plant size to find long-run minimizers. Looking forward: mastering cost functions enables analysis of production functions and duality (cost minimization), long-run competitive equilibria (entry/exit and zero-profit conditions), industrial organization topics like natural monopoly and oligopoly pricing strategies, and empirical estimation of cost and production technologies. Specific downstream concepts that require this material include: derivation of supply curves from cost functions, welfare analysis comparing price and marginal cost, and investment/capital choice models in dynamic firm theory.

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