Deriving the Competitive Firm’s Short-Run Supply

From the Output Decision to a Supply Rule

A competitive firm takes the market price as given. Its short-run supply rule describes the quantity it chooses at each possible price, holding its plant and other fixed commitments constant. The rule is not simply “produce where price equals marginal cost”: at sufficiently low prices, the firm may do better by temporarily producing nothing. Moreover, an equation such as price equals marginal cost identifies a candidate output; additional conditions are needed to show that the candidate is the firm’s global best choice.

Let $q\geq 0$ be output, $p$ the market price, $F$ unavoidable short-run fixed cost, and $VC(q)$ variable cost, with $VC(0)=0$. Assume $F$ is paid even when output is zero. The firm’s profit is $pq-VC(q)-F$. Since $F$ does not vary with $q$, choosing output is equivalent to maximizing the variable contribution $pq-VC(q)$. Average variable cost is $AVC(q)=VC(q)/q$ for $q>0$; marginal cost is $MC(q)=VC'(q)$ where the derivative exists.

Definition: The short-run shutdown threshold is $p_s=\inf_{q>0} AVC(q)$. It is the greatest lower bound of average variable cost over positive output, and need not be attained. At a price below $p_s$, every positive output has price below average variable cost, so producing adds a negative variable contribution. If the infimum is attained, producing at a minimizing quantity ties with shutdown when $p=p_s$. If it is not attained, every positive output has average variable cost strictly above $p_s$, so zero output is optimal at the threshold too.

The infimum, rather than an unconditional minimum, matters. For example, if $AVC(q)=3+\tfrac12q$, then $p_s=3$, but there is no positive quantity with $AVC(q)=3$. At price 3, producing any positive quantity loses money relative to producing zero, before fixed cost is counted.

Why the Shutdown Threshold Meets Marginal Cost

For a differentiable variable-cost function, the connection between average and marginal cost follows from differentiation. At a positive interior minimum of average variable cost, the derivative is zero. The calculation also explains why the intersection is relevant specifically when the minimum is attained at positive output.

AVC(q)=\frac{VC(q)}{q}, \qquad AVC'(q)=\frac{qVC'(q)-VC(q)}{q^2} =\frac{MC(q)-AVC(q)}{q}.

If $AVC$ has a differentiable minimum at $q_s>0$, then $AVC'(q_s)=0$. Since $q_s>0$, the displayed identity gives $MC(q_s)=AVC(q_s)=p_s$. This is a local calculus result; by itself it does not prove that the associated output is globally optimal for every price.

Derivation: At an interior average-variable-cost minimum, $AVC'(q_s)=0$. Substituting into $AVC'(q)=(MC(q)-AVC(q))/q$ yields $MC(q_s)-AVC(q_s)=0$. Thus marginal cost equals average variable cost at that minimum. If the infimum occurs only as $q$ approaches zero, there is no positive $q_s$ at which this argument can be applied.

For a price-taking firm, the derivative of variable contribution is $p-MC(q)$. An interior output that maximizes contribution therefore satisfies $p=MC(q)$. But this first-order condition alone is not a supply rule: there may be multiple solutions, a solution may be a local minimum, or a solution may yield less profit than shutdown. A reliable supply result needs both a global maximum and a reason the relevant rising-cost solution is globally best.

Conditions for the Short-Run Supply Result

The standard textbook result applies when the cost curve has a single relevant minimum of average variable cost and the rising marginal-cost branch gives the global best positive output. More precisely, suppose the firm’s variable contribution has a global maximum for each price considered; $AVC$ has a unique attained minimum at $q_s>0$; $MC(q_s)=AVC(q_s)=p_s$; and, for every $p>p_s$, a global maximizing positive output lies on the branch $q>q_s$, where $MC$ is strictly increasing and eventually exceeds $p$. These conditions rule out an unattained optimum at ever-higher output and prevent a competing lower-output branch from being better.

Under those conditions, for $p<p_s$, every positive output has $p-AVC(q)<0$, so shutdown is best. At $p=p_s$, zero output and $q_s$ both have zero variable contribution, so they tie. For $p>p_s$, some positive output has $AVC(q)<p$, and hence positive variable contribution. The global maximizing output is on the rising marginal-cost branch and satisfies $p=MC(q)$.

Theorem: Under the existence and global-branch conditions just stated, the firm’s short-run supply is zero below $p_s$, is $q_s$ or zero (both optimal) at $p_s$, and for $p>p_s$ is the unique solution $q>q_s$ to $MC(q)=p$. In the usual graph, this is the portion of the rising marginal-cost curve at or above minimum average variable cost, together with zero output below shutdown.

A common sufficient condition for existence is that $VC$ is continuous and grows faster than linearly: $VC(q)/q$ tends to infinity as $q$ tends to infinity. Then, for each finite price, $pq-VC(q)$ tends to negative infinity, so a continuous variable-contribution function attains a maximum on $q\geq0$. This existence condition does not, on its own, establish that a particular rising-$MC$ solution is globally optimal; the branch condition must also be checked. In other cost models, global convexity of $VC$ can make the marginal-cost first-order condition globally sufficient, but with $VC(0)=0$ a convex variable-cost function has nondecreasing average variable cost, so its shutdown infimum may be approached as $q$ tends to zero rather than attained at a positive quantity.

Worked Applications

Worked Example: Finding Output Above an Attained Shutdown Threshold

A price-taking workshop has $VC(q)=q^3-4q^2+8q$ for $q\geq0$, and unavoidable fixed cost $F=\$6$. Assume this cost schedule applies at all feasible outputs. Find its shutdown threshold, then find its optimal output and profit when price is $\$11$.

Method. Compute $AVC$ and $MC$, verify the threshold and rising branch, then check the global output choice. Since $VC(q)/q=q^2-4q+8=(q-2)^2+4$, the unique minimum is $p_s=\$4$ at $q_s=2$. The marginal-cost schedule is $MC(q)=3q^2-8q+8$, and $MC(2)=12-16+8=4$, as required. Its slope is $6q-8$, which is positive for $q>4/3$, including the branch $q\geq2$.

Set price equal to marginal cost: $11=3q^2-8q+8$, or $3q^2-8q-3=0$. Factoring gives $(3q+1)(q-3)=0$, so the nonnegative candidate is $q=3$, on the rising branch. To check the global choice, the derivative of variable contribution is $11-MC(q)$. Its only nonnegative zero is 3; it is positive below 3 and negative above 3. Thus contribution increases up to 3 and decreases afterward, making 3 the global maximizer.

At $q=3$, revenue is $11(3)=\$33$, variable cost is $27-36+24=\$15$, and variable contribution is $33-15=\$18$. Profit is $18-6=\$12$. Shutdown profit is $-\$6$, so producing is better. The result avoids the mistake of treating the fixed cost as part of average variable cost: $F$ affects profit but not the shutdown threshold.

Worked Example: An Infimum That Is Not Attained

A small supplier has $VC(q)=3q+\tfrac12q^2$ and unavoidable fixed cost $F=\$4$, with $q\geq0$. Determine the shutdown decision at price $\$3$, and the optimal output and profit at price $\$5$.

Here $AVC(q)=3+\tfrac12q$, so $p_s=\inf_{q>0}AVC(q)=\$3$. No positive quantity attains this infimum: every $q>0$ gives $AVC(q)>3$. At $p=3$, variable contribution is $3q-(3q+\tfrac12q^2)=-\tfrac12q^2<0$ for all $q>0$. Therefore the firm shuts down and earns $-\$4$; there is no positive-output tie at the threshold.

At $p=5$, $MC(q)=3+q$. Setting $MC=5$ gives $q=2$. Since variable contribution is $5q-3q-\tfrac12q^2=2q-\tfrac12q^2$, its derivative is $2-q$, which is positive for $q<2$ and negative for $q>2$. Thus $q=2$ is the global maximum. Revenue is $5(2)=\$10$; variable cost is $3(2)+\tfrac12(2^2)=6+2=\$8$; profit is $10-8-4=-\$2$. Shutdown would yield $-\$4$, so operating at a loss is preferable. The example shows why “minimum AVC” can be inaccurate when the infimum is not attained.

Worked Example: A Marginal-Cost Candidate Below Shutdown

Use $VC(q)=q^3-4q^2+8q$ and $F=\$9$, but now let price be $\$3$. A candidate from the rising marginal-cost branch is not automatically the firm’s supply. Decide whether the firm produces.

The shutdown threshold remains $\$4$, since $AVC(q)=(q-2)^2+4$. As price is below the threshold, for every positive $q$, variable contribution equals $q[3-AVC(q)]<0$. Therefore zero output is globally better than any positive output.

The tempting marginal-cost calculation is $3=3q^2-8q+8$, which gives $3q^2-8q+5=0$, or $(3q-5)(q-1)=0$. The root $q=5/3$ lies on the rising branch. Yet $AVC(5/3)=(1/3)^2+4=37/9$, so variable contribution there is $(5/3)(3-37/9)=(5/3)(-10/9)=-50/27$. Including fixed cost, profit is $-9-50/27=-293/27$, approximately $-\$10.85$, below shutdown profit $-\$9$. The correct choice is $q=0$. The trap is using a $p=MC$ solution without first applying the shutdown test.

Worked Example: Indifference at an Attained Threshold

A firm has $VC(q)=q^3-4q^2+8q$, unavoidable fixed cost $F=\$6$, and price $p=\$4$. Compare shutdown with producing the quantity that minimizes average variable cost.

As calculated above, $AVC(q)=(q-2)^2+4$, so the minimum is attained at $q_s=2$ and equals the price. At zero output, variable contribution is zero and profit is $0-6=-\$6$. At $q=2$, revenue is $4(2)=\$8$, variable cost is $8-16+16=\$8$, and variable contribution is also zero. Profit is again $-\$6$.

For any other positive quantity, $AVC(q)>4$, so $4q-VC(q)=q[4-AVC(q)]<0$. Thus no other output is better, and both $q=0$ and $q=2$ maximize profit. At the attained shutdown threshold, the firm is indifferent between shutting down and producing the minimizing quantity.

Common Mistakes and Economic Use

The shutdown rule separates the firm’s short-run operating choice from its fixed-cost burden. An unavoidable fixed cost is incurred whether output is positive or zero; it cancels when comparing those choices. That is why minimum average variable cost, not average total cost, sets the short-run shutdown threshold. Average total cost is relevant to whether revenue covers all costs, but a firm may rationally operate while making an economic loss if revenue covers variable cost and contributes toward fixed cost.

Using a minimum when only an infimum exists. Write $p_s=\inf_{q>0}AVC(q)$. If the infimum is not attained, there is no positive quantity tied with shutdown at the threshold.
Applying $p=MC$ before the shutdown test. A stationary point can have negative variable contribution. Compare it with the zero-output choice.
Assuming a local condition proves a global result. The identity $p=MC$ locates an interior candidate; it does not establish that the candidate exists, is a maximum, or beats every other feasible output.
Including fixed cost in $AVC$. Unavoidable fixed cost is not variable cost and should not enter the short-run shutdown threshold.

The supply curve is useful because it translates a price into a firm’s desired quantity, given its short-run cost conditions. It also makes explicit that below shutdown the firm supplies zero, while above shutdown the marginal-cost rule describes its positive-output choice only when the relevant branch is globally optimal.

Check Your Understanding

For each problem, identify the shutdown logic before using a marginal-cost equation.

A firm has $VC(q)=2q+q^2$, $F=\$5$, and price $p=\$6$. Find its shutdown threshold, optimal output, and profit. (Answer: $p_s=\$2$, $q=2$, profit $-\$1$.)
A firm has $AVC(q)=(q-3)^2+2$, with unavoidable fixed cost $F=\$7$, and price $p=\$2$. Identify all optimal outputs and the profit. (Answer: $q=0$ or $q=3$, both optimal; profit $-\$7$.)
A firm has $VC(q)=q^3-4q^2+8q$, $F=\$9$, and $p=\$3$. Solve $p=MC(q)$ for nonnegative candidates, then decide whether either candidate should be produced. (Answer: candidates $q=1$ and $q=5/3$; optimal output $q=0$, profit $-\$9$.)
A firm has $VC(q)=q^3-2q^2+6q$, $F=\$3$, and $p=\$9$. Its $AVC$ is $(q-1)^2+5$. Find the shutdown threshold and the optimal positive output. (Answer: $p_s=\$5$, $q=(2+\sqrt{13})/3$.)
A firm’s average variable cost is $AVC(q)=4+q$ for $q>0$, and its price is $\$4$. Is there a positive quantity that ties with shutdown? Explain using the threshold definition. (Answer: No; $p_s=\$4$ is an unattained infimum, and $q=0$ is the unique optimal output.)

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