The Long-Run Cost Curve as a Short-Run Envelope

Introduction

A firm with a fixed plant can adjust its output in the short run, but it cannot immediately change the plant itself. In the long run, it can choose a different plant design. The resulting long-run average-cost curve reflects the best available plant for each output level: it is the lower envelope of the short-run average-cost curves, not generally one particular short-run curve.

The word “best” needs care. A firm can compare the costs of all plant designs that can produce a given output. The lowest cost may be achieved by one of them, or it may only be approached without being attained. This distinction determines whether the long-run curve actually touches a short-run curve. The smooth tangency result requires still more conditions: a differentiable cost function and an attained, interior plant choice.

Short-Run Curves and the Long-Run Envelope

Definition: Let $q>0$ denote output and let $K$ denote a plant design or plant size. For a fixed plant $K$, the short-run total cost of producing $q$ is $C(q,K)$, and its short-run average cost is $SAC(q,K)=C(q,K)/q$. Let $\mathcal K(q)$ be the set of plant designs feasible for producing $q$. Long-run cost and long-run average cost are $$ C_{LR}(q)=\inf_{K\in\mathcal K(q)} C(q,K), \qquad LRAC(q)=\frac{C_{LR}(q)}{q}. $$

The cost function and feasible set are taken as given, with input prices and other relevant conditions held fixed. In the short run, the firm is restricted to one chosen plant $K$; in the long run, it may select among the feasible designs. A plant that cannot produce $q$ is not included in $\mathcal K(q)$. For $q>0$, dividing cost by $q$ preserves the ordering of costs, so the average-cost version can also be written as

LRAC(q)=\inf_{K\in\mathcal K(q)} SAC(q,K).

This is the lower-envelope relationship. For every feasible plant $K$, the infimum cannot exceed the cost associated with that particular plant:

LRAC(q)\leq SAC(q,K) \qquad\text{for every }K\in\mathcal K(q).

Thus the long-run curve lies at or below every feasible short-run average-cost curve. It follows the least-cost option as output changes. If the cost-minimizing plant changes with output, different short-run curves may form different parts of the envelope.

An infimum is the greatest lower bound of a set of values. It equals a minimum only if some feasible plant actually achieves that value. For example, if $\mathcal K(q)$ is compact and $C(q,K)$ is continuous in $K$, the minimum is attained. Those are sufficient conditions, not the only possible ones. If the feasible set is open or noncompact, or the cost function does not meet appropriate conditions, the infimum may not be attained. In that case the long-run curve is still defined, but there is no plant whose short-run curve touches it at that output.

Worked Example: Finding the Lower-Cost Plant at Three Output Levels

Method: Calculate both feasible short-run average costs at each output and select the lower value. This comparison avoids assuming that one plant is best at every output. A firm can choose either of two plant designs, $A$ and $B$. Their total costs, in thousands of dollars, are $C(q,A)=16+4q+q^2$ and $C(q,B)=4+8q+q^2$, for $q>0$. Calculate their short-run average costs and the long-run average cost at $q=2,3,4$.

Divide each total-cost function by $q$:

SAC(q,A)=\frac{16+4q+q^2}{q}=\frac{16}{q}+4+q, \qquad SAC(q,B)=\frac{4+8q+q^2}{q}=\frac{4}{q}+8+q.

At $q=2$, the two values are

SAC(2,A)=\frac{16}{2}+4+2=14, \qquad SAC(2,B)=\frac{4}{2}+8+2=12.

Therefore $LRAC(2)=\min(14,12)=12$, attained by plant $B$. At $q=3$,

SAC(3,A)=\frac{16}{3}+4+3=\frac{37}{3}, \qquad SAC(3,B)=\frac{4}{3}+8+3=\frac{37}{3}.

Both plants attain the same lowest value, so $LRAC(3)=37/3$, or $12\frac{1}{3}$ thousand dollars per unit of output. At $q=4$,

SAC(4,A)=\frac{16}{4}+4+4=12, \qquad SAC(4,B)=\frac{4}{4}+8+4=13.

Thus $LRAC(4)=12$, attained by plant $A$. The envelope switches from $B$ to $A$, with both plants equally economical at $q=3$. A common mistake is to choose the plant with the smaller fixed-cost term and assume it is always cheaper. The calculations show why the full average-cost expressions must be compared at each output. The tie at $q=3$ checks that the two curves cross there.

When Does the Envelope Touch a Short-Run Curve?

If a minimizing plant exists at output $q$, the long-run average cost equals the short-run average cost of that plant. That is what “touches” means in the envelope description: the curves have the same value at that output. A smooth tangency is a stronger claim. It says that, under appropriate differentiability and interior-choice conditions, the curves also have the same slope there.

To derive the slope condition, suppose the feasible plant set is the same around the output being considered and that $SAC(q,K)$ is differentiable. Assume that for each nearby $q$, the infimum is attained by an interior choice $K^*(q)$, with $K^*(q)$ differentiable. Then

LRAC(q)=SAC\bigl(q,K^*(q)\bigr).

Differentiate this equality using the chain rule. Because $K^*(q)$ minimizes short-run average cost at the chosen output, the first-order condition for an interior optimum is $\partial SAC(q,K^*)/\partial K=0$. Consequently, the term involving the change in plant size drops out:

\frac{d\,LRAC(q)}{dq} = \frac{\partial SAC(q,K^*(q))}{\partial q} + \frac{\partial SAC(q,K^*(q))}{\partial K}\frac{dK^*(q)}{dq} = \frac{\partial SAC(q,K^*(q))}{\partial q}.

The right side is the slope of the short-run average-cost curve for the chosen, fixed plant, evaluated at that output. The equality of slopes, along with equality of the curve values, gives the tangency interpretation. This argument is an application of the envelope principle: at the optimized plant choice, a small change in the choice variable has no first-order effect on the optimized value.

The conclusion is conditional. At a boundary plant choice, the first-order condition in $K$ need not hold. With a discrete set of plant designs, $K^*(q)$ may jump when the least-cost design changes. The long-run envelope can then have a kink at a crossing rather than a smooth tangency. If the infimum is not attained, there is no chosen plant at which to assert contact or tangency.

Worked Example: Checking a Smooth Tangency

Method: Minimize total cost over a continuous plant choice, then compare the long-run and fixed-plant average-cost slopes. This verifies tangency rather than inferring it merely from equal curve values. Suppose the cost of producing $q>0$ with plant size $K>0$ is $C(q,K)=q^2/K+K^2$. Find long-run average cost and check the tangency at $q=4$.

For fixed $q$, differentiate total cost with respect to $K$. The first-order condition is

\frac{\partial C}{\partial K}=-\frac{q^2}{K^2}+2K=0 \quad\Longrightarrow\quad q^2=2K^3 \quad\Longrightarrow\quad K^*(q)=\left(\frac{q^2}{2}\right)^{1/3}.

This is a minimum: the second derivative is $2q^2/K^3+2>0$ for $q,K>0$. Substituting the first-order condition into cost gives $q^2/K^*=2(K^*)^2$, so

C_{LR}(q)=3(K^*(q))^2 =3\left(\frac{q^2}{2}\right)^{2/3}, \qquad LRAC(q)=3\cdot 2^{-2/3}q^{1/3}.

At $q=4$, $K^*(4)=(16/2)^{1/3}=2$. Total cost is $4^2/2+2^2=8+4=12$, so average cost is $12/4=3$. The long-run formula gives $3\cdot 2^{-2/3}\cdot 4^{1/3}=3$, the same value.

For a fixed plant $K=2$, average cost as output varies is

SAC(q,2)=\frac{q^2/2+4}{q}=\frac{q}{2}+\frac{4}{q}.

Its slope at $q=4$ is $1/2-4/16=1/4$. The long-run slope is

\frac{d\,LRAC}{dq} = \frac{1}{3}\left(3\cdot2^{-2/3}q^{1/3}\right)\frac{1}{q} = \frac{1}{4} \qquad\text{at }q=4,

since $LRAC(4)=3$. Both curves have value $3$ and slope $1/4$ at the chosen output, confirming tangency. The calculation also illustrates the conditions behind the result: the minimizing $K=2$ is positive and interior, and the cost functions are differentiable.

Envelope Without Smooth Tangency

A lower envelope does not have to be a single smooth curve. The two-plant comparison above provides a discrete example. The derivatives of its short-run average costs are

\frac{d\,SAC(q,A)}{dq}=1-\frac{16}{q^2}, \qquad \frac{d\,SAC(q,B)}{dq}=1-\frac{4}{q^2}.

At the crossing $q=3$, these slopes are $-7/9$ for plant $A$ and $5/9$ for plant $B$. The long-run curve takes the lower branch on either side: plant $B$ below $q=3$, and plant $A$ above $q=3$. Its slope therefore changes at the crossing. Both curves attain the envelope value there, but neither has the same slope as the other, so the envelope has a kink rather than a smooth tangency.

This distinction is useful when interpreting diagrams and cost calculations. “The long-run curve is the lower envelope” describes its level relative to feasible short-run curves. “The curves are tangent” describes a local slope relationship and requires additional assumptions. In particular, an interior optimum in a smooth, continuously adjustable plant choice is not the same as a choice among a few fixed alternatives.

Why the Envelope Matters

The envelope captures the economic meaning of long-run flexibility. At a particular output, it records the lowest average cost available from feasible plant choices. As output changes, the plant that supports that lowest cost may change as well. Comparing the short-run curves therefore helps explain how a firm can face different average costs when it is temporarily locked into a plant than when it can redesign its production capacity.

A common exam trap is to treat the “best plant” as automatically existing and then write a minimum without qualification. For a general feasible set and cost function, the safe definition uses an infimum. If the problem states or establishes conditions ensuring an attained optimum, one may use a minimum and identify a cost-minimizing plant. A related trap is to infer tangency from the lower-envelope relationship alone. The envelope inequality guarantees that long-run average cost is no higher than any feasible short-run average cost; it does not by itself guarantee contact, differentiability, or equal slopes.

Key takeaway: Long-run average cost is the infimum of feasible short-run average costs at each output. If a minimizing plant is attained, the envelope meets that plant’s curve. If the optimum is also interior and the relevant functions are differentiable, the envelope and the selected short-run curve have equal slopes. Discrete choices or unattained infima need not produce smooth tangency.

This treatment is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Check Your Understanding

For each problem, distinguish the envelope comparison from any additional claim about attainment or tangency.

Two feasible plants have average-cost functions $SAC(q,A)=q+5+9/q$ and $SAC(q,B)=q+2+18/q$, for $q>0$. Find the output where their costs are equal, calculate the common average cost there, and identify the lower-cost plant at $q=1$ and $q=6$. (Answer: They are equal at $q=3$, where both costs are $11$; plant $A$ is lower at $q=1$, and plant $B$ is lower at $q=6$.)
A continuous plant choice $K>0$ has total cost $C(q,K)=q^2/K+4K$. Find the cost-minimizing $K$, derive $LRAC(q)$, and state the minimizing plant and average cost at $q=6$. (Answer: $K^*(q)=q/2$, $LRAC(q)=4$, and at $q=6$, $K^*=3$ with average cost $4$.)
For plants $A$ and $B$, let $SAC(q,A)=q+4+16/q$ and $SAC(q,B)=q+8+4/q$. Determine the crossing output and the slope of each curve there. Does the discrete envelope have a smooth tangency at the crossing? (Answer: They cross at $q=3$, where both equal $37/3$; the slopes are $-7/9$ and $5/9$, so the envelope has a kink, not a smooth tangency.)
Let $K\in(0,1)$ and suppose the total cost at output $q>0$ is $C(q,K)=K$. For a fixed output $q=2$, find the long-run cost and determine whether any feasible plant attains it. (Answer: $C_{LR}(2)=\inf_{0<K<1}K=0$; no feasible plant attains the infimum.)
Suppose $LRAC(q)=\inf_K SAC(q,K)$, and at some output $q_0$ the infimum is attained at an interior $K^*(q_0)$. State the first-order condition and, under differentiability and a differentiable optimal-choice function, state the resulting slope relationship. (Answer: $\partial SAC(q_0,K^*)/\partial K=0$, and $dLRAC/dq=\partial SAC(q,K^*(q))/\partial q$ at $q_0$; the selected short-run curve and long-run envelope have equal slopes there.)

What You'll Learn